<s>[INST] How find the minimum of the $|w^3+z^3|$,if $|z+w|=1,|z^2+w^2|=14$ let complex $z,w$ such
$$|z+w|=1,|z^2+w^2|=14$$
find the minimum of the value
$$|w^3+z^3|$$
My idea: let
$$z=a+bi,w=c+di\Longrightarrow z+w=(a+c)+(b+d)i,z^2+w^2=(a^2+b^2+c^2+d^2)+2(ab+cd)i$$
then we have
$$(a+c)^2+(b+d)^2=1,(a^2+b^2+c^2+d^2)+4(ab+cd)^2=14^2$$
and $$w^3+z^3=(a^3+c^3-3ab^2-3cd^2)+(3a^2b+3c^2d-b^3-d^3)i$$
 [/INST] Let $z+w = a$ and $z^2+w^2 = b$. Then, $zw = \dfrac{(z+w)^2-(z^2+w^2)}{2} = \dfrac{a^2-b}{2}$. 
Thus, $z^3+w^3 = (z+w)^3-3zw(z+w) = a^3 - 3 \cdot \dfrac{a^2-b}{2} \cdot a = \dfrac{3ab-a^3}{2}$. 
If $|a| = |z+w| = 1$ and $|b| = |z^2+w^2| = 14$, then using the reverse triangle inequality, we have 
$|z^3+w^3| = \left|\dfrac{3ab-a^3}{2}\right| = \dfrac{|a| \cdot |3b-a^2|}{2} = \dfrac{|3b-a^2|}{2} \ge \dfrac{3|b|-|a|^2}{2} = \dfrac{3\cdot 14 - 1^2}{2} = \dfrac{41}{2}$.
Equality occurs iff $3b$ and $a^2$ are $180^{\circ}$ apart when plotted in the complex plane. 
For instance, if $a = i$ and $b = 14$, then we get $z,w = \dfrac{\pm \sqrt{29} + i}{2}$ and $z^3+w^3 = \dfrac{41}{2}i$.
Therefore, the minimum value of $|z^3+w^3|$ is $\dfrac{41}{2}$. 
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