1
$\begingroup$

Prove that : $(a+b+c)^3-(b+c)^3-(c+a)^3-(a+b)^3+a^3+b^3+c^3=6abc$

Since $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a) $

Therefore the equation becomes : $2(a^3+b^3+c^3)+3(a+b)(b+c)(c+a) - [(c+a)^3 +(a+b)^3 +(b+c)^3]$

Putting $A:=(c+a)$ ; $B:=(a+b)$ ; $C:=(b+c)$

$[(c+a)^3 +(a+b)^3 +(b+c)^3] $ becomes $(A^3+B^3+C^3)$ now again using the formulae:$ a^3+b^3+c^3 = (a+b+c)^3-3(a+b)(b+c)(c+a)$

Could you please guide if any other easier way of doing this...

$\endgroup$
1
  • $\begingroup$ Please consider accepting some of the answers to your questions that you liked the best. For more information please refer to this meta question on "why should we accept answers?" $\endgroup$
    – Gigili
    Commented Mar 5, 2013 at 13:38

2 Answers 2

1
$\begingroup$

If you simply expand everything and collect terms, everything will cancel out and reduce down to 6abc. However, if you want to factorize it, think about the other terms you have -

As you said, $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)$. So then what does $(a+b)^3$ equal? If you expand it out following the same format as $(a+b+c)^3$ you get $a^3+b^3+3ab(a+b)$

So now your expression becomes $$a^3+b^3+c^3+3(a+b)(b+c)(c+a)-a^3-b^3-3ab(a+b)-b^3-c^3-3bc(b+c)-a^3-c^3-3ac(a+c)+a^3+b^3+c^3$$

Clean up all the cubes in there and you get $$3(a+b)(b+c)(c+a)-3ab(a+b)-3bc(b+c)-3ac(a+c)$$

Expand the first part of that to get $$3(2abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)-3ab(a+b)-3bc(b+c)-3ac(a+c)$$

Which becomes $$3(2abc+ab(a+b)+bc(b+c)+ac(a+c))-3(ab(a+b)+bc(b+c)+ac(a+c))$$

Now just cancel out all those like terms and you get $6abc$

$\endgroup$
3
  • $\begingroup$ thanks a lot for such a nice explanation...thanks a ton once again.. $\endgroup$ Commented Mar 5, 2013 at 13:16
  • $\begingroup$ @SachinSharmaa I think there is a shorter way to proving this equation.. Give me few minutes I would write it for you. $\endgroup$ Commented Mar 5, 2013 at 13:17
  • $\begingroup$ There is a beautiful symmetry but one hour has already past and I still can't figure out how to prove this equation more elegantly.. $\endgroup$ Commented Mar 5, 2013 at 14:13
1
$\begingroup$

OK - if you need a really shorter way (which may work depending on the equation you have):

You have $(a+b+c)^3-(b+c)^3-(c+a)^3-(a+b)^3+a^3+b^3+c^3=6abc$

Looking at the LHS as a cubic polynomial in $a$, we immediately can notice $a=0$ is a root and hence $a$ is a factor. By symmetry, $b$ and $c$ are factors. As $abc$ must be a factor of the LHS (which is homogeneous and also of degree $3$), the only other factor possible is a scalar. Hence LHS is of form $k\cdot abc$ and the scalar $k$ can be obtained by checking for easy values, say $a = b = 1$ and $c = -1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .