What is the value of (a+b)^3+(b+c)^3+(c+a)^3 for the roots of 8x^3+1001x+2008=0?

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The discussion focuses on calculating the expression \((a+b)^3+(b+c)^3+(c+a)^3\) for the roots \(a, b, c\) of the polynomial equation \(8x^3 + 1001x + 2008 = 0\). The roots can be derived using Vieta's formulas, which relate the coefficients of the polynomial to sums and products of its roots. The final value of the expression is determined to be \(3 \times (a+b+c)^3 - 3abc\), where \(a+b+c = -\frac{1001}{8}\) and \(abc = -\frac{2008}{8}\). This leads to a definitive numerical result based on the roots of the cubic equation.

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Let $a,\,b,\,c$ be the three roots of the equation $8x^3+1001x+2008=0$.

Find $(a+b)^3+(b+c)^3+(c+a)^3$.
 
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We have from vieta's formula
$a+b+ c = 0\cdots(1)$
$abc = - \frac{2008}{8}\cdots(2)$

hence a+b =-c
b+c = -a
c + a = -b

Hence $(a+b)^3 + (b+c)^3 + (c+a)^3 = -c^3-a^3-b^3 = - (c^3+a^3+b^3)\cdots(3)$

as $a+b+c=0$ so $a^3+b^+c^3 = 3abc \cdots(4)$

from (3) and (4)

$(a+b)^3 + (b+c)^3 + (c+a)^3 = - 3abc = -3 (-\frac{2008}{8})= -753 $ (using (2)

Hence Ans = - 753
 
kaliprasad said:
We have from vieta's formula
$a+b+ c = 0\cdots(1)$
$abc = - \frac{2008}{8}\cdots(2)$

hence a+b =-c
b+c = -a
c + a = -b

Hence $(a+b)^3 + (b+c)^3 + (c+a)^3 = -c^3-a^3-b^3 = - (c^3+a^3+b^3)\cdots(3)$

as $a+b+c=0$ so $a^3+b^+c^3 = 3abc \cdots(4)$

from (3) and (4)

$(a+b)^3 + (b+c)^3 + (c+a)^3 = - 3abc = -3 (-\frac{2008}{8})= -753 $ (using (2)

Hence Ans = - 753

there were issues and I could not edit the solution hence doing below

We have from vieta's formula
$a+b+ c = 0\cdots(1)$
$abc = - \frac{2008}{8}\cdots(2)$

hence a+b =-c
b+c = -a
c + a = -b

Hence $(a+b)^3 + (b+c)^3 + (c+a)^3 = -c^3-a^3-b^3 = - (c^3+a^3+b^3)\cdots(3)$

as $a+b+c=0$ so $a^3+b^3+c^3 = 3abc \cdots(4)$

from (3) and (4)

$(a+b)^3 + (b+c)^3 + (c+a)^3 = - 3abc = -3 (-\frac{2008}{8})= 753 $ (using (2)

Hence Ans = 753
 

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