If $a$ and $b$ both are positive integers then
$m = a+b, \;n = a^2 + b^2, \; p^3 = a^3+b^3$ Then show that $ m^3+2p^3 = 3mn$.
This is a easy problem to solve. Just substitute those values and show that.
But I can't find a way to solve this
$n = a^2 + b^2, \; p^3 = a^3+b^3, \; m^3 + 2p^3 = 3mn$ Then prove that $m = a+b$ is a solution to the system.
Any hint will be helpful.
On
$$m^3+2p^3=3mn\implies m^3+2(a^3+b^3)=3m(a^2+b^2)$$
$$\iff m^3-3m(a^2+b^2)+2(a^3+b^3)=0$$
Let $m=u+v\implies m^3=u^3+v^3+3uvm\iff m^3-3m(uv)-(u^3+v^3)=0$
Comparing we get $u^3+v^3=-2(a^3+b^3)$ and $uv=a^2+b^2=u^3v^3=(a^2+b^2)^3$
So, $u^3,v^3$ are the roots of $$t^2+2(a^3+b^3)t+(a^2+b^2)^3=0\ \ \ \ (1)$$
But $a^3,b^3$ are in general not the roots of $(1)$
You showed that if $m = a+b, \;n = a^2 + b^2, \; p^3 = a^3+b^3$ Then show that $ m^3+2p^3 = 3mn$.
Thus, imagine you are solving the cubic equation $x^3+2p^3=3xn$. Let one such solution be $x=m$.
From the upper statement, note that $m=a+b$ is a solution. Our proof is done.