Can anyone prove it

Prove that If a+b+c=0 and a,b,c are real numbers, then prove a^5/5+b^5/5+c^5/5 = (a^3/3+b^3/3+c^3/3)(a^2/2+b^2/2+c^2/2)

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Consider a,b,c to be the roots of the cubic equation x^3+sx+t=0, because a+b+c=0. So a^3+sa+t=0, b^3+sb+t=0, c^3+sc+t=0. So a^3+b^3+c^3+3t=0, by adding the above equations. So (a^3+b^3+c^3)= - 3t.

Also a+b+c=0, so a^2+b^2+c^2+2(ab+bc+ca)=0, (a^2^+b^2+c^2)= -(ab+bc+ca)= -2s.

Finally, x^3+sx+t=0 implies that x^5+sx^3+tx^2=0,

so, a^5+sa^3+ta^2=0, b^5+sb^3+tb^2=0, c^5+sc^3+tc^2=0.

adding these, we get (a^5+b^5+c^5)-3st-2st=0, So (a^5+b^5+c^5)/5 = st =(-s)(-t) = (a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/2

Hence, proved.

Shourya Pandey - 8 years, 3 months ago

Great,but I think there is a mistake ,in the starting of second line, it should be b^3 + sb +t=0, and not b^3+as+t=0....Then after adding those equations we will get -3t, otherwise we will get -3t-3as.

kiran patel - 8 years, 3 months ago

Changes made...

Shourya Pandey - 8 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$