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

@Shourya Pandey
–
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
·
4 years ago

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TopNewestConsider 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 · 4 years ago

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– Kiran Patel · 4 years 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.Log in to reply

– Shourya Pandey · 4 years ago

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