let f(a,b)=(a+b)^7a^7b^7
now f(0,b)=0 and f(a,0)=0 thus we have
f(a,b)=k*ab*g(a,b)
where k is a constant and g(a,b) is a polynomial in a,b
now if a=b then f(a,b)=0 as well thus we can simplify further to
f(a,b)=k*ab*(a+b)*g(a,b)
again with k a constant and g(a,b) a polynomial in a,b
now by expanding f(a,b) and dividing out ab(a+b) we can see that
k=7 and g(a,b)=a^4+2a^3b+3a^2b^2+2ab^3+b^4
now this is where a bit of intuition comes into play. The symmetry in the coefficients makes me think of squared polynomial thus we have
g(a,b)=[h(a,b)]^2
now we can deduce that h(a,b) is a polynomial in a,b of degree two with 3 terms two of which are a^2 and b^2 and since the remaining terms of g(a,b) have a common factor of ab then the third term of h(a,b) must be ab
thus we can check if
g(a,b)=(a^2+ab+b^2)^2
which by expanding is verified
thus we are left with
(a+b)^7a^7b^7=7ab(a+b)(a^2+ab+b^2)^2

Posted by Daniel
on 20150608 11:47:38 