Question 967458
Let {{{a/b = b/c = k}}}
therefore {{{b = ck}}} and {{{a = ck^2}}}
So, LHS = {{{(ck^2+ck+c)*(ck^2-ck+c)}}}
= {{{c*(k^2+k+1)*c*(k^2-k+1)}}}
= {{{c^2*((k^2+1)^2-k^2)}}} (Using {{{(x+y)(x-y)=x^2-y^2}}})
= {{{c^2*(k^4+1+2k^2-k^2)}}}
= {{{c^2*(k^4+k^2+1)}}}
= {{{c^2*k^4+c^2*k^2+c^2}}}
= {{{(ck^2)^2 + (ck)^2 + c^2}}}
= {{{a^2+b^2+c^2}}}
= RHS. proved.