Question

If $a+b+c=0$, show that $6(a^5+b^5+c^5)=5(a^3+b^3+c^3)(a^2+b^2+c^2)$.

Answer 1

We have identically
$(1+ax)(1+bx)(1+cx)=1+px+q{x}^2+r{x}^3$,
where $p=a+b+c$, $q=bc+ca+ab$, $r=abc$.
Hence, using the condition given,
$(1+ax)(1+bx)(1+cx)=1+q{x}^2+r{x}^3$.
Taking logarithms and equating the coefficients of $x^n$, we have $\frac{{(-1)}^{n-1}}{n}(a^n+b^n+c^n)=$ coefficient of $x^n$ in the expansion of $\log (1+q{x}^2+r{x}^3)=$ coefficient of $x^n$ in $(q{x}^2+r{x}^3)-\frac{1}{2}{(q{x}^2+r{x}^3)}^2+\frac{1}{3}{(q{x}^2+r{x}^3)}^3-\cdots$
By putting $n=2,3,5$ we obtain
$-\frac{a^2+b^2+c^2}{2}=q$, $\frac{a^3+b^3+c^3}{3}=r$, $\frac{a^5+b^5+c^5}{5}=-qr$;
whence $\frac{a^5+b^5+c^5}{5}=\frac{a^3+b^3+c^3}{3}\cdot \frac{a^2+b^2+c^2}{2}$,
and the required result at once follows.
If $a=\beta -\gamma ,b=\gamma -\alpha ,c=\alpha -\beta$, the given condition is satisfied; hence we have identically for all values of $\alpha ,\beta ,\gamma$
$6\{{(\beta -\gamma )}^5+{(\gamma -\alpha )}^5+{(\alpha -\beta )}^5\}=5\{{(\beta -\gamma )}^3+{(\gamma -\alpha )}^3+{(\alpha -\beta )}^3\}\{{(\beta -\gamma )}^2+{(\gamma -\alpha )}^2+{(\alpha -\beta )}^2\}$
that is,
$6\{{(\beta -\gamma )}^5+{(\gamma -\alpha )}^5+{(\alpha -\beta )}^5\}=5(\beta -\gamma )(\gamma -\alpha )(\alpha -\beta )({\alpha }^2+{\beta }^2+{\gamma }^2-\beta \gamma -\gamma \alpha -\alpha \beta )$;

$6(a^5+b^5+c^5)=5\left(a^3+b^3+c^3\right)(a^2+b^2+c^2)$

Hence proved.