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Solve the following equations:

Suppose x1, x2 and x3 are roots of ${(11 - x)}^{3}$ + ${(13 - x)}^{3}$ = ${(24 - 2 x)}^{3}$ . What is the sum of x1 + x2 + x3 ?

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-36

-34

answer is A.

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Detailed Solution

Given:
The roots of

{(11 - x)}^{3}

{(13 - x)}^{3}

{(24 - 2 x)}^{3}

are x1, x2 and x3 .
Let us consider a general cubic equation of the form a

x^{3}

+ b

x^{2}

+ cx + d = 0, where a, b, c, d is non-zero. So, let us consider p, q, and r are the roots. Then, we can write the relationships as below,
p + q + r = −

\frac{b}{a}

pq + qr + rp =

\frac{b}{a}

pqr = −

\frac{d}{a}

−
For

{(11 - x)}^{3}

, the expansion is,
⇒

{(11 - x)}^{3}

11^{3}

– 3 ×

11^{2}

× x + 3 × 11 ×

x^{2}

−

x^{3}

[

{(a - b)}^{3}

a^{3}

− 3

a^{2}

b + 3a

b^{2}

−

b^{3}

]
⇒

{(11 - x)}^{3}

= 1331 − 363x + 33

x^{2}

−

x^{3}

…..…..….. (1)
For

{(13 - x)}^{3}

the expansion is,
⇒

{(13 - x)}^{3}

13^{3}

– 3 ×

13^{2}

× x + 3 × 13 ×

x^{2}

−

x^{3}

⇒

{(13 - x)}^{3}

= 2197 – 507x + 39

x^{2}

−

x^{3}

…..…..….. (2)
For

{(24 - 2 x)}^{3}

, the expansion is,
⇒

{(24 - 2 x)}^{3}

24^{3}

– 3 ×

24^{2}

× 2x + 3 × 24 ×

{(2 x)}^{2}

−

{(2 x)}^{3}

⇒

{(24 - 2 x)}^{3}

= 13824 − 3456x + 288

x^{2}

− 8

x^{3}

…..…..….. (3)
The equation is,

{(11 - x)}^{3}

{(13 - x)}^{3}

{(24 - 2 x)}^{3}

Now, substitute the values from equation (1), (2), and (3)
⇒ (1331 − 363x + 33

x^{2}

−

x^{3}

) + (2197 – 507x + 39

x^{2}

−

x^{3}

) = (13824 − 3456x + 288

x^{2}

− 8

x^{3}

)
⇒ 3528 − 870x + 72

x^{2}

− 2

x^{3}

= 13824 − 3456x + 288

x^{2}

− 8

x^{3}

⇒ (3528 − 870x + 72

x^{2}

− 2

x^{3}

) − (13824 − 3456x + 288

x^{2}

− 8

x^{3}

) = 0
⇒ 6

x^{3}

− 216

x^{2}

+ 2586x – 10296 = 0
So, using the relationships in the cubic equation given to us, we can write sum of products as,
p + q + r = −

\frac{b}{a}

Substitute the values,
⇒ x1 + x2 + x3 = −

\frac{(- 216)}{6}

∴ x1 + x2 + x3 = −36
Hence, the value of x1 + x2 + x3 is -36.
So, option (1) is correct.

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Solve the following equations:

Suppose x1, x2 and x3 are roots of (11 - x)3 + (13 - x)3 = (24 - 2x)3. What is the sum of x1 + x2 + x3 ?

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Suppose x1, x2 and x3 are roots of ${(11 - x)}^{3}$ + ${(13 - x)}^{3}$ = ${(24 - 2 x)}^{3}$ . What is the sum of x1 + x2 + x3 ?