Question

If $a+b+c=0$, prove that identities $a^6+b^6+c^6=3a^2{b}^2{c}^2-2(bc+ca+ab{)}^3$.

Answer 1

$\Rightarrow (1+ax)(1+bx)(1+cx)=1+px+q{x}^2+r{x}^3$

$p=a+b+c=0$

$q=\sum ab$

$r=abc$

Taking the logarithms and equating the co efficients of $x^n$ we have $\frac{{(-1)}^n}{n}(a^n+b^n+c^n)$

=co efficient of $x^n$ in the expansion of $\log (1+q{x}^2+r{x}^3)$

=co efficient of $x^n$ in $(q{x}^2+r{x}^3)$

$-\frac{1}{2}{(q{x}^2+r{x}^3)}^2$

$+\frac{1}{3}{(q{x}^2+r{x}^3)}^3+.....$

Putting, $n=2,3,4,5,6,7$

$-\frac{a^2+b^2+c^2}{2}=q,\frac{a^3+b^3+c^3}{3}=r,$

$-\frac{a^4+b^4+c^4}{4}=-\frac{q^2}{2},\frac{a^5+b^5+c^5}{5}=-qr,$

$-\frac{a^6+b^6+c^6}{6}=-\frac{r^2}{2}+\frac{q^3}{3},\frac{a^7+b^7+c^7}{7}=q^{2}r,$

$LHS$;-

$a^6+b^6+c^6=3r^2-2q^3$

$RHS$;-

$3a^2{b}^2{c}^2-2{(bc+ab+ca)}^3=3r^2-2q^3$