If $z_1^2+z_2^2$ is real, $z_1(z_1^2-3z_2^2)=2$, and $z_2(3z_1^2-z_2^2)=11$, then find $(z_1^2+z_2^2)^2$ 
If $z_1$ and $z_2$ are complex numbers such that $z_1^2+z_2^2 \in\mathbb R$ and $$z_1(z_1^2-3z_2^2)=2,\qquad z_2(3z_1^2-z_2^2)=11,$$ then find the value of $(z_1^2+z_2^2)^2$. Given answer is $25$.

I have tried many things but I am not getting the answer.
I subtracted two equations to observe that $11/z_2 - 2/z_1$ must be real. Also if $z_1=x_1+iy_1$ and $z_2=x_2+i y_2$, then using the fact that $z_1^2+z_2^2 \in\mathbb R$, we get $x_1 y_1+x_2y_2=0$ but I am not able to compile these results to get the desired value.

I suggest that we square both equations.
$$z_1^2(z_1^2-3z_2^2)^2=4\\
z_2^2(3z_1^2-z_2^2)^2=121$$
Now add them together and simplify.
$$125=z_1^2(z_1^2-3z_2^2)^2+z_2^2(3z_1^2-z_2^2)^2=z_1^2(z_1^4-6z_1^2z_2^2+9z_2^4)+z_2^2(9z_1^4-6z_1^2z_2^2+z_2^4)=\\
=z_1^6+3z_1^4z_2^2+3z_1^2z_2^4+z_2^6=(z_1^2+z_2^2)^3$$
The rest is trivial.
