MathematicsFactorize a3b-c3+b3c-a3+c3a-b3.

Factorize a3b-c3+b3c-a3+c3a-b3.


  1. A
    3abc(b-c)(c+a)(a-b)
  2. B
    3abc(b-c)(c-a)(a-b)
  3. C
    3abc(b-c)(c-a)(a+b)  
  4. D
    3abc(b-c)(c-a)(b-a)  

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    Solution:

    Given, a3b-c3+b3c-a3+c3a-b3.
    We know the formula,
    x3+y3+z3-3xyz=(x+y+z)x2+y2+z2-xy-yz-zx .
    If (x+y+z)=0, then x3+y3+z3=3xyz.
    Therefore, rewrite the given expression we have,
    a3(b-c)3+b3(c-a)3+c3(a-b)3
    =[a(b-c)]3+[b(c-a)]3+[c(a-b)]3
    =(ab-ac)3+(bc-ab)3+(ac-bc)3
    Here, we have x=ab-ac, y=bc-ab, and z=ac-bc.
    Therefore, adding x, y and z, we get,
    x+y+z=(ab-ac)+(bc-ab)+(ac-bc)
    ab-ac+bc-ab+ac-bc=0
    a3(b-c)3+b3(c-a)3+c3(a-b)3
    (ab-ac)3+(bc-ab)3+(ac-bc)3
    3(ab-ac)(bc-ab)(ac-bc)
    3[a(b-c)][b(c-a)][c(a-b)]
    3abc(b-c)(c-a)(a-b)
    Hence, the required answer 3abc(b-c)(c-a)(a-b).
    Therefore, option 2 is correct.
     
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