Welcome to our community

Be a part of something great, join today!

[SOLVED] Find (a+b)³+(b+c)³+(c+a)³

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,588
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

Well-known member
Mar 31, 2013
1,279
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

Well-known member
Mar 31, 2013
1,279
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