Prove if $(a+b)$ divides $(a^2+ab+b^2)$ then $(a+b)^2$ divides $(a^4+b^4)$?

Question

I solved this problem for only $(a+b)$ divides $(a^4+b^4)$.

$(a^2+ab+b^2)=k(a+b)$

$\color{red}{(a^2-ab+b^2)}(a+b)^2=a^4+a^3b+ab^3+b^4$
$\color{orange}{(a^2+ab+b^2)}(a-b)^2=a^4-a^3b-ab^3+b^4$
$2(a^4+b^4)= \color{red}{m}(a+b)^2 + \color{orange}{p}(a-b)^2 =(a+b)(m(a+b)+ k(a-b)^2)$

Does anybody have an idea how to prove that $(a+b)^2$ divides $(a^4+b^4)$?

Answer 1

Hint: $a+b$ divides $(a^2 + ab + b^2) - b (a+b) = a^2$, and similarly divides $b^2$.