Study MaterialsRD Sharma SolutionsRD Sharma Solutions Class 9 Chapter 4 Algebraic Identities Ex 4.1

RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities Ex 4.1

RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities

RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities Ex 4.1

Question 1.
Evaluate each of the following using identities:
(i) (2x –\(\frac { 1 }{ x }\))2
(ii) (2x + y) (2x – y)
(iii) (a2b – b2a)2
(iv) (a – 0.1) (a + 0.1)
(v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2)
Solution:
RD Sharma Class 9 Chapter 4 Algebraic Identities
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities

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    Question 2.
    Evaluate each of the following using identities:
    (i) (399)2
    (ii) (0.98)2
    (iii) 991 x 1009
    (iv) 117 x 83
    Solution:
    RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities

    Question 3.
    Simplify each of the following:
    RD Sharma Class 9 PDF Chapter 4 Algebraic Identities
    Algebraic Identities Class 9 RD Sharma Solutions

    Solution:
    Question 4.
    Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities
    Solution:
    Class 9 Maths Chapter 4 Algebraic Identities RD Sharma Solutions
    RD Sharma Book Class 9 PDF Free Download Chapter 4 Algebraic Identities

    Question 5.
    RD Sharma Book Class 9 Pdf Free Download Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Class 9 Book Chapter 4 Algebraic Identities

    Question 6.
    Algebraic Identities Problems With Solutions PDF RD Sharma Class 9 Solutions
    Solution:
    RD Sharma Class 9 Maths Book Questions Chapter 4 Algebraic Identities

    Question 7.
    If 9x2 + 25y2 = 181 and xy = -6, find the value of 3x + 5y.
    Solution:
    9x2 + 25y2 = 181, and xy = -6
    (3x + 5y)2 = (3x)2 + (5y)2 + 2 x 3x + 5y
    ⇒ 9X2 + 25y2 + 30xy
    = 181 + 30 x (-6)
    = 181 – 180 = 1
    = (±1 )2
    ∴ 3x + 5y = ±1

    Question 8.
    If 2x + 3y = 8 and xy = 2, find the value of 4X2 + 9y2.
    Solution:
    2x + 3y = 8 and xy = 2
    Now, (2x + 3y)2 = (2x)2 + (3y)2 + 2 x 2x x 3y
    ⇒ (8)2 = 4x2 + 9y2 + 12xy
    ⇒ 64 = 4X2 + 9y2 + 12 x 2
    ⇒ 64 = 4x2 + 9y2 + 24
    ⇒ 4x2 + 9y2 = 64 – 24 = 40
    ∴ 4x2 + 9y2 = 40

    Question 9.
    If 3x -7y = 10 and xy = -1, find the value of 9x2 + 49y2
    Solution:
    3x – 7y = 10, xy = -1
    3x -7y= 10
    Squaring both sides,
    (3x – 7y)2 = (10)2
    ⇒ (3x)2 + (7y)2 – 2 x 3x x 7y = 100
    ⇒ 9X2 + 49y2 – 42xy = 100
    ⇒ 9x2 + 49y2 – 42(-l) = 100
    ⇒ 9x2 + 49y2 + 42 = 100
    ∴ 9x2 + 49y2 = 100 – 42 = 58

    Question 10.
    Simplify each of the following products:
    RD Sharma Mathematics Class 9 Solutions Chapter 4 Algebraic Identities
    Solution:
    Solution Of Rd Sharma Class 9 Chapter 4 Algebraic Identities
    RD Sharma Math Solution Class 9 Chapter 4 Algebraic Identities

    Question 11.
    RD Sharma Class 9 Questions Chapter 4 Algebraic Identities
    Solution:
    Maths RD Sharma Class 9 Chapter 4 Algebraic Identities

    Question 12.
    RD Sharma Class 9 Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities

    Question 13.
    Simplify each of the following products:
    RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Class 9 PDF Chapter 4 Algebraic Identities
    Algebraic Identities Class 9 RD Sharma Solutions

    Question 14.
    Prove that a2 + b2 + c2 – ab – bc – ca is always non-negative for all values of a, b and c.
    Solution:
    RD Sharma Class 9 Solution Chapter 4 Algebraic Identities
    ∵ The given expression is sum of these squares
    ∴ Its value is always positive Hence the given expression is always non-­negative for all values of a, b and c

    Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities Ex 4.2

    Question 1.
    Write the following in the expanded form:
    RD Sharma Class 9 Chapter 4 Algebraic Identities Ex 4.2
    Solution:
    RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities
    RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities

    Question 2.
    If a + b + c = 0 and a2 + b2 + c2 = 16, find the value of ab + be + ca.
    Solution:
    a + b+ c = 0
    Squaring both sides,
    (a + b + c)2 = 0
    ⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 0
    16 + 2(ab + bc + c) = 0
    ⇒ 2(ab + bc + ca) = -16
    ⇒ ab + bc + ca =-\(\frac { 16 }{ 2 }\) = -8
    ∴ ab + bc + ca = -8

    Question 3.
    If a2 + b2 + c2 = 16 and ab + bc + ca = 10, find the value of a + b + c.
    Solution:
    (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
    = 16 + 2 x 10
    = 16 + 20 = 36
    = (±6)2
    ∴ a + b + c = ±6

    Question 4.
    If a + b + c = 9 and ab + bc + ca = 23, find the value of a2 + b2 + c2.
    Solution:
    (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
    ⇒ (9)2 = a2 + b2 + c2 + 2 x 23
    81= a2 + b2 + c2 + 46
    a2 + b2 + c2 = 81 – 46 = 35
    a2 + b2 + c2 = 35

    Question 5.
    Find the value of 4x2 + y2 + 25z2 + 4xy – 10yz – 20zx when x = 4, y = 3 and z = 2.
    Solution:
    x = 4, y – 3, z = 2
    4x2 + y2 + 25z2 + 4xy – 10yz – 20zx
    = (2x)2 + (y)2 + (5z)2 + 2 x2 x x y-2 x y x 5z – 2 x 5z x 2x
    = (2x + y- 5z)2
    = (2 x 4 + 3- 5 x 2)2
    = (8 + 3- 10)2
    = (11 – 10)2
    = (1)2 = 1

    Question 6.
    Simplify:
    (i) (a + b + c)2 + (a – b + c)2
    (ii) (a + b + c)2 – (a – b + c)2
    (iii) (a + b + c)2 + (a – b + c)2 + (a + b – c)2
    (iv) (2x + p – c)2 – (2x – p + c)2
    (v) (x2 + y2 – z2)2 – (x2 – y2 + z2)2
    Solution:

    RD Sharma Class 9 PDF Chapter 4 Algebraic Identities
    Algebraic Identities Class 9 RD Sharma Solutions

    Question 7.
    Simplify each of the following expressions:
    Algebraic Identities Class 9 RD Sharma Solutions
    Solution:
    Class 9 Maths Chapter 4 Algebraic Identities RD Sharma Solutions
    Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities
    RD Sharma Book Class 9 Pdf Free Download Chapter 4 Algebraic Identities

    Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities Ex 4.3

    Question 1.
    Find the cube of each of the following binomial expressions:
    Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities
    Solution:
    Class 9 Maths Chapter 4 Algebraic Identities RD Sharma Solutions
    RD Sharma Book Class 9 PDF Free Download Chapter 4 Algebraic Identities
    RD Sharma Book Class 9 Pdf Free Download Chapter 4 Algebraic Identities

    Question 2.
    If a + b = 10 and ab = 21, find the value of a3 + b3.
    Solution:
    a + b = 10, ab = 21
    Cubing both sides,
    (a + b)3 = (10)3
    ⇒ a3 + 63 + 3ab (a + b) = 1000
    ⇒ a3 + b3 + 3 x 21 x 10 = 1000
    ⇒ a3 + b3 + 630 = 1000
    ⇒ a3 + b3 = 1000 – 630 = 370
    ∴ a3 + b3 = 370

    Question 3.
    If a – b = 4 and ab = 21, find the value of a3-b3.
    Solution:
    a – b = 4, ab= 21
    Cubing both sides,
    ⇒ (a – A)3 = (4)3
    ⇒ a3 – b3 – 3ab (a – b) = 64
    ⇒ a3-i3-3×21 x4 = 64
    ⇒ a3 – 63 – 252 = 64
    ⇒ a3 – 63 = 64 + 252 =316
    ∴ a3 – b3 = 316

    Question 4.
    RD Sharma Class 9 Book Chapter 4 Algebraic Identities
    Solution:
    Algebraic Identities Problems With Solutions PDF RD Sharma Class 9 Solutions
    RD Sharma Class 9 Maths Book Questions Chapter 4 Algebraic Identities

    Question 5.
    RD Sharma Mathematics Class 9 Solutions Chapter 4 Algebraic Identities
    Solution:
    Solution Of Rd Sharma Class 9 Chapter 4 Algebraic Identities

    Question 6.
    RD Sharma Math Solution Class 9 Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Class 9 Questions Chapter 4 Algebraic Identities
    Maths RD Sharma Class 9 Chapter 4 Algebraic Identities

    Question 7.
    RD Sharma Class 9 Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities

    Question 8.
    RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Class 9 PDF Chapter 4 Algebraic Identities
    Algebraic Identities Class 9 RD Sharma Solutions

    Question 9.
    If 2x + 3y = 13 and xy = 6, find the value of 8x3 + 21y3.
    Solution:
    2x + 3y = 13, xy = 6
    Cubing both sides,
    (2x + 3y)3 = (13)3
    ⇒ (2x)3 + (3y)3 + 3 x 2x x 3X2x + 3y) = 2197
    ⇒ 8x3 + 27y3 + 18xy(2x + 3y) = 2197
    ⇒ 8x3 + 27y3 + 18 x 6 x 13 = 2197
    ⇒ 8X3 + 27y3 + 1404 = 2197
    ⇒ 8x3 + 27y3 = 2197 – 1404 = 793
    ∴ 8x3 + 27y3 = 793

    Question 10.
    If 3x – 2y= 11 and xy = 12, find the value of 27x3 – 8y3.
    Solution:
    3x – 2y = 11 and xy = 12 Cubing both sides,
    (3x – 2y)3 = (11)3
    ⇒ (3x)3 – (2y)3 – 3 x 3x x 2y(3x – 2y) =1331
    ⇒ 27x3 – 8y3 – 18xy(3x -2y) =1331
    ⇒ 27x3 – 8y3 – 18 x 12 x 11 = 1331
    ⇒ 27x3 – 8y3 – 2376 = 1331
    ⇒ 27X3 – 8y3 = 1331 + 2376 = 3707
    ∴ 2x3 – 8y3 = 3707

    Question 11.
    Evaluate each of the following:
    (i) (103)3
    (ii) (98)3
    (iii) (9.9)3
    (iv) (10.4)3
    (v) (598)3
    (vi) (99)3
    Solution:
    We know that (a + bf = a3 + b3 + 3ab(a + b) and (a – b)3= a3 – b3 – 3 ab(a – b)
    Therefore,
    (i) (103)3 = (100 + 3)3
    = (100)3 + (3)3 + 3 x 100 x 3(100 + 3) {∵ (a + b)3 = a3 + b3 + 3ab(a + b)}
    = 1000000 + 27 + 900 x 103
    = 1000000 + 27 + 92700
    = 1092727
    (ii) (98)3 = (100 – 2)3
    = (100)3 – (2)3 – 3 x 100 x 2(100 – 2)
    = 1000000 – 8 – 600 x 98
    = 1000000 – 8 – 58800
    = 1000000-58808
    = 941192
    (iii) (9.9)3 = (10 – 0.1)3
    = (10)3 – (0.1)3 – 3 X 10 X 0.1(10 – 0.1)
    = 1000 – 0.001 – 3 x 9.9
    = 1000 – 0.001 – 29.7
    = 1000 – 29.701
    = 970.299
    (iv) (10.4)3 = (10 + 0.4)3
    = (10)3 + (0.4)3 + 3 x 10 x 0.4(10 + 0.4)
    = 1000 + 0.064 + 12(10.4)
    = 1000 + 0.064 + 124.8 = 1124.864
    (v) (598)3 = (600 – 2)3
    = (600)3 – (2)3 – 3 x 600 x 2 x (600 – 2)
    = 216000000 – 8 – 3600 x 598
    = 216000000 – 8 – 2152800
    = 216000000 – 2152808
    = 213847192
    (vi) (99)3 = (100 – 1)3
    = (100)3 – (1)3 – 3 x 100 x 1 x (100 – 1)
    = 1000000 – 1 – 300 x 99
    = 1000000 – 1 – 29700
    = 1000000 – 29701
    = 970299

    Question 12.
    Evaluate each of the following:
    (i) 1113 – 893
    (ii) 463 + 343
    (iii) 1043 + 963
    (iv) 933 – 1073
    Solution:
    We know that a3 + b3 = (a + bf – 3ab(a + b) and a3 – b3 = (a – bf + 3 ab(a – b)
    (i) 1113 – 893
    = (111 – 89)3 + 3 x ill x 89(111 – 89)
    = (22)3 + 3 x 111 x 89 x 22
    = 10648 + 652014 = 662662
    (OR)
    (a + b)3 – (a – b)3 = 2(b3 + 3a2b)
    = 1113 – 893 = (100 + 11)3 – (100 – 11)3
    = 2(113 + 3 x 1002 x 11] = 2(1331 + 330000] = 331331 x 2 = 662662
    (a + b)3 + (a- b)3 = 2(b3 + 3ab2)
    (ii) 463 + 343 = (40 + 6)3 + (40 – 6)3
    = 2[(40)3 + 3 x 40 x 62] = 2[64000 + 3 x 40 x 36] = 2[64000 + 4320] = 2 x 68320 = 136640
    (iii) 1043 + 963 = (100 + 4)3 + (100 – 96)3
    = 2 [a3 + 3 ab2] = 2[(100)3 + 3 x 100 x (4)2] = 2[ 1000000 + 300 x 16] = 2[ 1000000 + 4800] = 1004800 x 2 = 2009600
    (iv) 933 – 1073 = -[(107)3 – (93)3] = -[(100 + If – (100 – 7)3] = -2[b3 + 3a2b)] = -2[(7)3 + 3(100)2 x 7] = -2(343 + 3 x 10000 x 7] = -2[343 + 210000] = -2[210343] = -420686

    Question 13.
    RD Sharma Class 9 Solution Chapter 4 Algebraic Identities
    Solution:
    Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities
    Class 9 Maths Chapter 4 Algebraic Identities RD Sharma Solutions

    Question 14.
    Find the value of 27X3 + 8y3 if
    (i) 3x + 2y = 14 and xy = 8
    (ii) 3x + 2y = 20 and xy = \(\frac { 14 }{ 9 }\)
    Solution:
    RD Sharma Book Class 9 PDF Free Download Chapter 4 Algebraic Identities
    RD Sharma Class 9 Book Chapter 4 Algebraic Identities

    Question 15.
    Find the value of 64x3 – 125z3, if 4x – 5z = 16 and xz = 12.
    Solution:
    4x – 5z = 16, xz = 12
    Cubing both sides,
    (4x – 5z)3 = (16)3
    ⇒ (4x)3 – (5y)3 – 3 x 4x x 5z(4x – 5z) = 4096
    ⇒ 64x3 – 125z3 – 3 x 4 x 5 x xz(4x – 5z) = 4096
    ⇒ 64x3 – 125z3 – 60 x 12 x 16 = 4096
    ⇒ 64x3 – 125z3 – 11520 = 4096
    ⇒ 64x3 – 125z3 = 4096 + 11520 = 15616

    Question 16.
    Algebraic Identities Problems With Solutions PDF RD Sharma Class 9 Solutions
    Solution:
    RD Sharma Class 9 Maths Book Questions Chapter 4 Algebraic Identities
    RD Sharma Mathematics Class 9 Solutions Chapter 4 Algebraic Identities

    Question 17.
    Simplify each of the following:
    Solution Of Rd Sharma Class 9 Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Math Solution Class 9 Chapter 4 Algebraic Identities
    RD Sharma Class 9 Questions Chapter 4 Algebraic Identities
    Maths RD Sharma Class 9 Chapter 4 Algebraic Identities

    Question 18.
    RD Sharma Class 9 Chapter 4 Algebraic Identities
    RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities
    RD Sharma Class 9 PDF Chapter 4 Algebraic Identities

    Question 19.
    Algebraic Identities Class 9 RD Sharma Solutions
    Solution:
    RD Sharma Class 9 Solution Chapter 4 Algebraic Identities
    Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities

    RD Sharma Mathematics Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.4

    Question 1.
    Find the following products:
    (i) (3x + 2y) (9X2 – 6xy + Ay2)
    (ii) (4x – 5y) (16x2 + 20xy + 25y2)
    (iii) (7p4 + q) (49p8 – 7p4q + q2)
    RD Sharma Class 9 Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities
    RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities
    RD Sharma Class 9 PDF Chapter 4 Algebraic Identities
    Algebraic Identities Class 9 RD Sharma Solutions

    Question 2.
    If x = 3 and y = -1, find the values of each of the following using in identity:
    RD Sharma Class 9 Solution Chapter 4 Algebraic Identities
    Solution:
    Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities
    Class 9 Maths Chapter 4 Algebraic Identities RD Sharma Solutions
    RD Sharma Book Class 9 PDF Free Download Chapter 4 Algebraic Identities
    RD Sharma Class 9 Book Chapter 4 Algebraic Identities

    Question 3.
    If a + b = 10 and ab = 16, find the value of a2 – ab + b2 and a2 + ab + b2.
    Solution:
    a + b = 10, ab = 16 Squaring,
    (a + b)2 = (10)2
    ⇒ a2 + b2 + lab = 100
    ⇒ a2 + b2 + 2 x 16 = 100
    ⇒ a2 + b2 + 32 = 100
    ∴ a2 + b2 = 100 – 32 = 68
    Now, a2 – ab + b2 = a2 + b2 – ab = 68 – 16 = 52
    and a2 + ab + b2 = a2 + b2 + ab = 68 + 16 = 84

    Question 4.
    If a + b = 8 and ab = 6, find the value of a3 + b3.
    Solution:
    a + b = 8, ab = 6
    Cubing both sides,
    (a + b)3 = (8)3
    ⇒ a3 + b3 + 3 ab{a + b) = 512
    ⇒ a3 + b3 + 3 x 6 x 8 = 512
    ⇒ a3 + b3 + 144 = 512
    ⇒ a3 + b3 = 512 – 144 = 368
    ∴ a3 + b3 = 368

    Question 5.
    If a – b = 6 and ab = 20, find the value of a3-b3.
    Solution:
    a – b = 6, ab = 20
    Cubing both sides,
    (a – b)3 = (6)3
    ⇒ a3 – b3 – 3ab(a – b) = 216
    ⇒ a3 – b3 – 3 x 20 x 6 = 216
    ⇒ a3 – b3 – 360 = 216
    ⇒ a3 -b3 = 216 + 360 = 576
    ∴ a3 – b3 = 576

    Question 6.
    If x = -2 and y = 1, by using an identity find the value of the following:
    Algebraic Identities Problems With Solutions PDF RD Sharma Class 9 Solutions
    Solution:
    RD Sharma Class 9 Maths Book Questions Chapter 4 Algebraic Identities

    RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities Ex 4.5

    Question 1.
    Find the following products:
    (i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx)
    (ii) (4x -3y + 2z) (16x2 + 9y2+ 4z2 + 12xy + 6yz – 8zx)
    (iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca)
    (iv) (3x -4y + 5z) (9x2 + 16y2 + 25z2 + 12xy- 15zx + 20yz)
    Solution:
    (i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx)
    = (3x + 2y + 2z) [(3x)2 + (2y)2 + (2z)2 – 3x x 2y + 2y x 2z + 2z x 3x] = (3x)3 + (2y)3 + (2z)3 – 3 x 3x x 2y x 2z
    = 27x3 + 8y3 + 8Z3 – 36xyz
    (ii) (4x – 3y + 2z) (16x2 + 9y2 + 4z2 + 12xy + 6yz – 8zx)
    = (4x -3y + 2z) [(4x)2 + (-3y)2 + (2z)2 – 4x x (-3y + (3y) x (2z) – (2z x 4x)] = (4x)3 + (-3y)3 + (2z)3 – 3 x 4x x (-3y) x (2z)
    = 64x3 – 27y3 + 8z3 + 72xyz
    (iii) (2a -3b- 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca)
    = (2a -3b- 2c) [(2a)2 + (3b)2 + (2c)2 – 2a x (-3b) – (-3b) x (-2c) – (-2c) x 2a] = (2a)3 + (3b)3 + (-2c)3 -3x 2a x (-3 b) (-2c)
    = 8a3 – 21b3 -8c3 – 36abc
    (iv) (3x – 4y + 5z) (9x2 + 16y2 + 25z2 + 12xy – 15zx + 20yz)
    = [3x + (-4y) + 5z] [(3x)2 + (-4y)2 + (5z)2 – 3x x (-4y) -(-4y) (5z) – 5z x 3x] = (3x)3 + (-4y)3 + (5z)3 – 3 x 3x x (-4y) (5z)
    = 27x3 – 64y3 + 125z3 + 180xyz

    Question 2.
    Evaluate:
    Solution Of Rd Sharma Class 9 Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Class 9 Chapter 4 Algebraic Identities
    RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities

    Question 3.
    If x + y + z = 8 and xy + yz+ zx = 20, find the value of x3 + y3 + z3 – 3xyz.
    Solution:
    We know that
    x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 -xy -yz – zx)
    Now, x + y + z = 8
    Squaring, we get
    (x + y + z)2 = (8)2
    x2 + y2 + z2 + 2(xy + yz + zx) = 64
    ⇒ x2 + y2 + z2 + 2 x 20 = 64
    ⇒ x2 + y2 + z2 + 40 = 64
    ⇒ x2 + y2 + z2 = 64 – 40 = 24
    Now,
    x3 + y3 + z3 – 3xyz = (x + y + z) [x2 + y2 + z2 – (xy + yz + zx)] = 8(24 – 20) = 8 x 4 = 32

    Question 4.
    If a +b + c = 9 and ab + bc + ca = 26, find the value of a3 + b3 + c3 – 3abc.
    Solution:
    a + b + c = 9, ab + be + ca = 26
    Squaring, we get
    (a + b + c)2 = (9)2
    a2 + b2 + c2 + 2 (ab + be + ca) = 81
    ⇒ a2 + b2 + c2 + 2 x 26 = 81
    ⇒ a2 + b2 + c2 + 52 = 81
    ∴ a2 + b2 + c2 = 81 – 52 = 29
    Now, a3 + b3 + c3 – 3abc = (a + b + c) [(a2 + b2 + c2 – (ab + bc + ca)] = 9[29 – 26] = 9 x 3 = 27

    Question 5.
    If a + b + c = 9, and a2 + b2 + c2 = 35, find the value of a3 + b3 + c3 – 3abc.
    Solution:
    a + b + c = 9
    Squaring, we get
    (a + b + c)2 = (9)2
    ⇒ a2 + b2 + c2 + 2 (ab + be + ca) = 81
    ⇒ 35 + 2(ab + bc + ca) = 81
    2(ab + bc + ca) = 81 – 35 = 46
    ∴ ab + bc + ca = \(\frac { 46 }{ 2 }\) = 23
    Now, a3 + b3 + c3 – 3abc
    = (a + b + c) [a2 + b2 + c2 – (ab + bc + ca)] = 9[35 – 23] = 9 x 12 = 108

    Class 9 Maths Chapter 4 Algebraic Identities RD Sharma Solutions VSAQS

    Question 1.
    RD Sharma Class 9 PDF Chapter 4 Algebraic Identities
    Solution:
    Algebraic Identities Class 9 RD Sharma Solutions

    Question 2.
    RD Sharma Class 9 Solution Chapter 4 Algebraic Identities
    Solution:
    Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities
    Class 9 Maths Chapter 4 Algebraic Identities RD Sharma Solutions

    Question 3.
    If a + b = 7 and ab = 12, find the value of a2 + b2.
    Solution:
    a + b = 7, ab = 12
    Squaring both sides,
    (a + b)2 = (7)2
    ⇒ a2 + b2 + 2ab = 49
    ⇒ a2 + b2 + 2 x 12 = 49
    ⇒ a2 + b2 + 24 = 49
    ⇒ a2 + b2 = 49 – 24 = 25
    ∴ a2 + b2 = 25

    Question 4.
    If a – b = 5 and ab = 12, find the value of a2 + b2 .
    Solution:
    a – b = 5, ab = 12
    Squaring both sides,
    ⇒ (a – b)2 = (5)2
    ⇒ a2 + b2 – 2ab = 25
    ⇒ a2 + b2 – 2 x 12 = 25
    ⇒ a2 + b2 – 24 = 25
    ⇒ a2 + b2 = 25 + 24 = 49
    ∴ a2 + b2 = 49

    Question 5.
    RD Sharma Book Class 9 PDF Free Download Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Class 9 Book Chapter 4 Algebraic Identities

    Question 6.
    Algebraic Identities Problems With Solutions PDF RD Sharma Class 9 Solutions
    Solution:
    RD Sharma Class 9 Maths Book Questions Chapter 4 Algebraic Identities

    Question 7.
    RD Sharma Mathematics Class 9 Solutions Chapter 4 Algebraic Identities
    Solution:
    Solution Of Rd Sharma Class 9 Chapter 4 Algebraic Identities
    RD Sharma Math Solution Class 9 Chapter 4 Algebraic Identities

    Algebraic Identities Class 9 RD Sharma Solutions MCQS

    Question 1.
    Class 9 Maths Chapter 4 Algebraic Identities RD Sharma Solutions
    Solution:

    RD Sharma Book Class 9 PDF Free Download Chapter 4 Algebraic Identities

    Question 2.
    RD Sharma Class 9 Book Chapter 4 Algebraic Identities
    Solution:
    Algebraic Identities Problems With Solutions PDF RD Sharma Class 9 Solutions

    Question 3.
    RD Sharma Class 9 Maths Book Questions Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Mathematics Class 9 Solutions Chapter 4 Algebraic Identities

    Question 4.
    Solution Of Rd Sharma Class 9 Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Math Solution Class 9 Chapter 4 Algebraic Identities
    RD Sharma Class 9 Questions Chapter 4 Algebraic Identities

    Question 5.
    Maths RD Sharma Class 9 Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Class 9 Chapter 4 Algebraic Identities

    Question 6.
    RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities
    RD Sharma Class 9 PDF Chapter 4 Algebraic Identities

    Question 7.
    Algebraic Identities Class 9 RD Sharma Solutions
    Solution:
    RD Sharma Class 9 Solution Chapter 4 Algebraic Identities

    Question 8.
    If a + b + c = 9 and ab + bc + ca = 23, then a2 + b2 + c2 =
    (a) 35
    (b) 58
    (c) 127
    (d) none of these
    Solution:
    a + b + c = 9, ab + bc + ca = 23
    Squaring,
    (a + b+ c) = (9)2
    a2 + b2 + c2 + 2 (ab + bc + ca) = 81
    ⇒ a2 + b2 + c2 + 2 x 23 = 81
    ⇒ a2 + b2+ c2 + 46 = 81
    ⇒ a2 + b2+ c2 = 81 – 46 = 35 (a)

    Question 9.
    (a – b)3 + (b – c)3 + (c – a)3 =
    (a) (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
    (d) (a -b)(b- c) (c – a)
    (c) 3(a – b) (b – c) (c – a)
    (d) none of these
    Solution:
    (a – b)3 + (b- c)3 + (c- a)3
    ∵ a – b + b – c + c – a = 0
    ∴ (ab)3 + (b – c)3 + (c – a)3
    = 3
    (a -b)(b- c) (c – a) (c)

    Question 10.
    Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities
    Solution:
    Class 9 Maths Chapter 4 Algebraic Identities RD Sharma Solutions

    Question 11.
    If a – b = -8 and ab = -12 then a3 – b3 =
    (a) -244
    (b) -240
    (c) -224
    (d) -260
    Solution:
    a – b = -8, ab = -12
    (a – b)3 = a3 – b3 – 3ab (a – b)
    (-8)3 = a3 – b3 – 3 x (-12) (-8)
    -512 = a3-b3– 288
    a3 – b3 = -512 + 288 = -224 (c)

    Question 12.
    If the volume of a cuboid is 3x2 – 27, then its possible dimensions are
    (a) 3, x2, -27x
    (b) 3, x – 3, x + 3
    (c) 3, x2, 27x
    (d) 3, 3, 3
    Solution:
    Volume = 3x2 -27 = 3(x2 – 9)
    = 3(x + 3) (x – 3)
    ∴ Dimensions are = 3, x – 3, x + 3 (b)

    Question 13.
    75 x 75 + 2 x 75 x 25 + 25 x 25 is equal to
    (a) 10000
    (b) 6250
    (c) 7500
    (d) 3750
    Solution:
    RD Sharma Book Class 9 PDF Free Download Chapter 4 Algebraic Identities
    RD Sharma Class 9 Book Chapter 4 Algebraic Identities

    Question 14.
    (x – y) (x + y)(x2 + y2) (x4 + y4) is equal to
    (a) x16 – y16
    (b) x8 – y8
    (c) x8 + y8
    (d) x16 + y16
    Solution:
    Algebraic Identities Problems With Solutions PDF RD Sharma Class 9 Solutions

    Question 15.
    RD Sharma Class 9 Maths Book Questions Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Mathematics Class 9 Solutions Chapter 4 Algebraic Identities

    Question 16.
    Solution Of Rd Sharma Class 9 Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Math Solution Class 9 Chapter 4 Algebraic Identities
    RD Sharma Class 9 Questions Chapter 4 Algebraic Identities

    Question 17.
    RD Sharma Class 9 Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities

    Question 18.
    RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Class 9 PDF Chapter 4 Algebraic Identities
    Algebraic Identities Class 9 RD Sharma Solutions

    Question 19.
    If a2 + b2 + c2 – ab – bc – ca = 0, then
    (a) a + b = c
    (b) b + c = a
    (c) c + a = b
    (d) a = b = c
    Solution:
    a2 + b2 + c2 – ab – bc – ca = 0
    2 {a2 + b2 + c2 – ab – be – ca) = 0 (Multiplying by 2)
    ⇒ 2a2 + 2b2 + 2c2– 2ab – 2bc – 2ca = 0
    ⇒ a2 + b2 – 2ab + b2 + c2 – 2bc + c2 + a2 – 2ca = 0
    ⇒ (a – b)2 + (b – c)2 + (c – a)2 = 0
    (a – b)2 = 0, then a – b = 0
    ⇒ a = b
    Similarly, (b – c)2 = 0, then
    b-c = 0
    ⇒ b = c
    and (c – a)2 = 0, then c-a = 0
    ⇒ c = a
    ∴ a = b – c (d)

    Question 20.
    RD Sharma Class 9 Solution Chapter 4 Algebraic Identities
    Solution:
    Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities

    Question 21.
    Class 9 Maths Chapter 4 Algebraic Identities RD Sharma Solutions
    RD Sharma Book Class 9 PDF Free Download Chapter 4 Algebraic Identities
    Solution:
    Algebraic Identities Problems With Solutions PDF RD Sharma Class 9 Solutions

    Question 22.
    If a + b + c = 9 and ab + bc + ca = 23, then a3 + b3 + c3 – 3 abc =
    (a) 108
    (b) 207
    (c) 669
    (d) 729
    Solution:
    a3 + b3 + c3 – 3abc
    = (a + b + c) [a2 + b2 + c2 – (ab + bc + ca)
    Now, a + b + c = 9
    Squaring,
    a2 + b2 + c2 + 2 (ab + be + ca) = 81
    ⇒ a2 + b2 + c2 + 2 x 23 = 81
    ⇒ a2 + b2 + c2 + 46 = 81
    ⇒ a2 + b2 + c2 = 81 – 46 = 35
    Now, a3 + b3 + c3 – 3 abc = (a + b + c) [(a2 + b2 + c2) – (ab + bc + ca)
    = 9[35 -23] = 9 x 12= 108 (a)

    Question 23.
    RD Sharma Class 9 Maths Book Questions Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Mathematics Class 9 Solutions Chapter 4 Algebraic Identities
    Solution Of Rd Sharma Class 9 Chapter 4 Algebraic Identities

    Question 24.
    The product (a + b) (a – b) (a2 – ab + b2) (a2 + ab + b2) is equal to
    (a) a6 + b6
    (b) a6 – b6
    (c) a3 – b3
    (d) a3 + b3
    Solution:
    (a + b) (a – b) (a2 – ab + b2) (a2 + ab +b2)
    = (a + b) (a2-ab + b2) (a-b) (a2 + ab + b2)
    = (a3 + b3) (a3 – b3)
    = (a3)2 – (b3)2 = a6 – b6 (b)

    Question 25.
    The product (x2 – 1) (x4 + x2 + 1) is equal to
    (a) x8 – 1
    (b) x8 + 1
    (c) x6 – 1
    (d) x6 + 1
    Solution:
    (x2 – 1) (x4 + x2 + 1)
    = (x2)3 – (1)3 = x6 – 1 (c)

    Question 26.
    RD Sharma Math Solution Class 9 Chapter 4 Algebraic Identities
    Solution:
    RD Sharma Class 9 Questions Chapter 4 Algebraic Identities

    Question 27.
    Maths RD Sharma Class 9 Chapter 4 Algebraic Identities
    Solution:
    Maths RD Sharma Class 9 Chapter 4 Algebraic Identities

     

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