If $z_1$ and $z_2$ are two complex numbers,and if $z_1^3-3z_1^2z_2=2,3z_1z_2^2-z_2^3=11$, If $z_1$ and $z_2$ are two complex numbers,and if $z_1^3-3z_1^2z_2=2,3z_1z_2^2-z_2^3=11$,then find the value of $|z_1^2+z_2^2|$.

$z_1^3-3z_1^2z_2=2$
$3z_1z_2^2-z_2^3=11$
Adding them,we get
$z_1^3-3z_1^2z_2+3z_1z_2^2-z_2^3=13$
$(z_1-z_2)^3=13$
We need to find $|z_1^2+z_2^2|$,
I could not solve from here,I am stuck.Please help me.Thanks.
 A: dividing both the equations we get
$$\frac{z_1^2}{z_2^2} \times \frac{z_1-3z_2}{3z_1-z_2}=\frac{2}{11}$$ $\implies$ assuming $\frac{z_1}{z_2}=p$ we get
$$\frac{p^2(p-3)}{3p-1}=\frac{2}{11}$$ i.e.,
$$11p^3-33p^2-6p+2=0$$ and this cubic equation has three real roots which means
$\frac{z_1}{z_2}$ is Real.
Now $$|z_1^2+z_2^2|=|p^2z_2^2+z_2^2|=|z_2^2|(p^2+1) \tag{1}$$
also $$(z_1-z_2)^3=13$$ $\implies$
$$(pz_2-z_2)^3=13$$, so
$$z_2^3=\frac{13}{(p-1)^3}$$ so
$$|z_2|^2=\frac{13^{\frac{2}{3}}}{(p-1)^2}$$ substituting this in $(1)$ we get
$$|z_1^2+z_2^2|=\frac{13^{\frac{2}{3}}(p^2+1)}{(p-1)^2}$$ so we have three answers for each $p \in \mathbb{R}$
A: Hint:Assume $z_1=x+iy,z_2=a+bi$ . You now have two equations . Substitute the above values in place of $z_1,z_2$ you will get the desired result by some algebra.As you observe since numbers are real which implies imaginary parts of both equations is $0$ so each equation got two parts. So you get 4 equations and 4 unknowns by comparing real and imaginary part Also a more help. $(z_1^2+z_2^2)=\frac{2z_2+11z_1}{z_1z_2}$. Thats enough for solving it.
A: $$z_{1}(z^2_{1}-3z^2_{2}) = 2\Rightarrow z^3_{1}-3z_{1}z^2_{2} = 2\cdots \cdots (1)$$
Similarly $$z_{2}(3z^2_{1}-z_{2}^2) = 11\Rightarrow 3z^2_{1}z_{2}-z^3_{2} = 11\cdots (2)\times i$$
Now Add  and Subtract these two equation, we get 
$$(z_{1}+iz_{2})^3 = 2+11i$$
$$(z_{1}-iz_{2})^3 = 2-11i$$
So $$(z^2_{1}+z^2_{2})^3 = (2+11i)(2-11i) = 125\Rightarrow |z^2_{1}+z^2_{2}|=5$$
