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

[anemone]

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

Find $(a+b)^3+(b+c)^3+(c+a)^3$.

 

[kaliprasad]

Spoiler: my solution

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]

Spoiler: my solution

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

Spoiler: modfied solution

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