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- Feb 14, 2012

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

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,588

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

- Mar 31, 2013

- 1,279

$a+b+ c = 0\cdots(1)$

$abc = - \frac{2008}{8}\cdots(2)$

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

- Mar 31, 2013

- 1,279

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

$a+b+ c = 0\cdots(1)$

$abc = - \frac{2008}{8}\cdots(2)$

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

$a+b+ c = 0\cdots(1)$

$abc = - \frac{2008}{8}\cdots(2)$

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