Show that if $a+b+c=0$ then $\frac{a^2+b^2+c^2}{2} * \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5}$

Question

Title says it all. It's my homework and I don't know where to start. A good hint would be appreciated.

solve $a+b+c=0$ for $c$ and plug this in the left-Hand side and the right-Hand side of your equation — Dr. Sonnhard Graubner, Nov 17 '15 at 18:52

score 5 · Accepted Answer · answered Nov 17 '15 at 18:56

5

Let $p,q$ be the numbers such that the roots of $x^3+px+q=0$ are $a,b,c$.

Then, by Vieta's formulas, $$a+b+c=0,\qquad ab+bc+ca=p\tag1$$

Now express $a^2+b^2+c^2,a^3+b^3+c^3$ and $a^5+b^5+c^5$ by $p,q$ using $(1)$ and $$x^3=-px-q,\qquad x^5=x^2(-px-q)=-px^3-qx^2.$$

answered Nov 17 '15 at 18:56

mathlove

I figured it out, thanks a lot! I really like this solution. – Mislav Blažević Nov 17 '15 at 20:35

score 3 · Answer 2 · answered Nov 17 '15 at 18:54

Replace $c$ by $-(a+b)$ and expand, then we can get following

$LHS=\frac{a^2+b^2+(a+b)^2}{2}*\frac{a^3+b^3-(a+b)^3}{3}=(a^2+b^2+ab)(a^2b+ab^2)=ab(a+b)(a^2+b^2+ab)$

$RHS=\frac{a^5+b^5-(a+b)^5}{5}=a^4b+2a^3b^2+2a^2b^3+ab^4=ab(a^3+b^3+2a^2b+2ab^2)=ab(a+b)(a^2+b^2-ab+2ab)=ab(a+b)(a^2+b^2+ab)$

Thus, you can see $LHS=RHS$

2 Answers2