## Monday, February 29, 2016

### Prove that $(a^3+b^3+c^3+d^3)^2=9(ab-cd)(bc-ad)(ca-bd)$.

Let $a,\,b,\,c,\,d$ be real numbers such that $a+b+c+d=0$.

Prove that $(a^3+b^3+c^3+d^3)^2=9(ab-cd)(bc-ad)(ca-bd)$.

My solution:

First, we draw some very useful and helpful identities from the given equality,  $a+b+c+d=0$:

$a+b+c+d=0$

[MATH]\color{yellow}\bbox[5px,purple]{a+b=-(c+d)}[/MATH]

$(a+b)^2=(-(c+d))^2$

$a^2+b^2+2ab=c^2+d^2+2cd$

[MATH]\implies \color{yellow}\bbox[5px,green]{c^2+d^2=2(ab-cd)+a^2+b^2}[/MATH]

[MATH]\implies \color{yellow}\bbox[5px,blue]{a^2+b^2-c^2-d^2+2ab-2cd=0}[/MATH]

[MATH]\implies \color{black}\bbox[5px,orange]{a^2+b^2-c^2-d^2=2cd-2ab}[/MATH]

Also, note that $a^3+b^3=(a+b)(a^2-ab+b^2)$.

We can now try to work on the LHS of the target equation to prove it equivalent to what it has on the RHS of the equation:

$(a^3+b^3+c^3+d^3)^2$

$=((a+b)(a^2-ab+b^2)+(c+d)(c^2-cd+d^2))^2$

$=((a+b)(a^2-ab+b^2)-(a+b)(c^2-cd+d^2))^2$ (since [MATH]\color{yellow}\bbox[5px,purple]{a+b=-(c+d)}[/MATH])

$=((a+b)(a^2-ab+b^2-c^2+cd-d^2))^2$

$=((a+b)(2cd-2ab-ab+cd))^2$ [MATH]\color{black}\bbox[5px,orange]{a^2+b^2-c^2-d^2=2cd-2ab}[/MATH]

$=(a+b)^2(3cd-3ab)^2$

$=3^2(a+b)^2(cd-ab)^2$

$=9(a+b)^2(ab-cd)^2$

$=9(ab-cd)(a+b)^2(ab-cd)$---(*)

The remaining effort now is to show that $(a+b)^2(ab-cd)=(bc-ad)(ac-bd)$. If we can prove that, we're then done.

That shouldn't be too hard a thing to do, and we can prove it two ways, we can either prove it backwards ($(bc-ad)(ac-bd)=(a+b)^2(ab-cd)$) or forwards ($(a+b)^2(ab-cd)=(bc-ad)(ac-bd)$),  as long as we know how to algebraically manipulate with what we've established:

$(bc-ad)(ac-bd)$

$=bcac-bcbd-adac+adbd$

$=abc^2-b^2cd-a^2cd+abd^2$

$=ab(c^2+d^2)-cd(a^2+b^2)$

$=ab(2(ab-cd)+a^2+b^2)-cd(a^2+b^2+2a-2ab)$ [MATH] \color{yellow}\bbox[5px,green]{c^2+d^2=2(ab-cd)+a^2+b^2}[/MATH]

$=ab(a^2+b^2+2ab-2cd)-cd(a^2+b^2+2a-2ab)$

$=ab(a^2+b^2+2ab)-2abcd-cd(a^2+b^2+2ab)+2abcd$

$=ab(a+b)^2-cd(a+b)^2$

$=(ab-cd)(a+b)^2$--(**)

Putting back what we have found in (**) back to (*), we see that we have proved $(a^3+b^3+c^3+d^3)^2=9(ab-cd)(bc-ad)(ca-bd)$, provided $a+b+c+d=0$.