Thread:HMcCoy/@comment-25501845-20151005153622/@comment-25501845-20151026153804

Finally, got it But, it was not Completely figured out by me, I seeked the help of my Math Teacher too.

Here is the answer, and thanks for Considering the Question :)

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Note that g(1) = f(1) = 0, so 1 is a root of both f(x) and g(x).

Let p and q be the other two roots of f(x), so p^2 and q^2 are the other two roots of g(x).

We then get

pq = −c and

p^2 q^2 = −a,

so ‘a’ = −c ^2.

Also,

(−a)^2 = (p+q + 1)^2 = p ^2 +q ^2 + 1 + 2(pq +p+q) = −b+ 2b = b.

Therefore

b = c ^4.

Since f(1) = 0

we therefore get 1 + c − c ^2 + c ^4 = 0.

Factorising, we get (c + 1)(c^ 3 − c ^2 + 1) = 0.

Note that c ^3 – c^ 2 + 1 = 0 has no integer root and hence

c = −1, b = 1, a = −1.

Therefore a^ 2013 + b^ 2013 + c ^2013 = −1.

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<p class="MsoNormal">I should really learn Latex and " Math " Tag!