Question

If $a+b+c=0$, prove the identity $a^5+b^5+c^5=-5abc(bc+ca+ab)$.

Answer 1

Assuming that $a,b\text{and}c$ are roots of a cubic equation, we can write

$(x-a)(x-b)(x-c)=0\text{or}{x}^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0$

$⟹x^3+(ab+bc+ca)x-abc=0[\because a+b+c=0]$

Also, as $a,b,c$ are the roots, they will satisfy the cubic equation. Hence,

$a^3+(ab+bc+ca)a-abc=0\dots \dots \dots \dots \dots \dots \dots \left(1\right)$

$b^3+(ab+bc+ca)b-abc=0\dots \dots \dots \dots \dots \dots \dots \left(2\right)$

$c^3+(ab+bc+ca)c-abc=0\dots \dots \dots \dots \dots \dots \dots \left(3\right)$

Adding the above equations, we get $a^3+b^3+c^3=3abc[\because a+b+c=0]$

Also, $(a^2+b^2+c^2)=(a+b+c{)}^2-2(ab+bc+ca)=-2(ab+bc+ca)$

Multiplying equation (1), (2), (3) with $a^2,b^2,c^2$ respectively and adding them, we get

$(a^5+b^5+c^5)+(ab+bc+ca)(a^3+b^3+c^3)-abc(a^2+b^2+c^2)=0$

Substituting value of $(a^2+b^2+c^2)$ and $(a^3+b^3+c^3)$ from earlier, the above equation becomes

$(a^5+b^5+c^5)+(ab+bc+ca)\left(3abc\right)+\left(2abc\right)(ab+bc+ca)=0$

$⟹a^5+b^5+c^5=-5abc(ab+bc+ca)$