<s>[INST] How to calculate $4 \over {{x^4} + {y^4} + {z^4}}$ from $x + y + z = 1$ and other conditions more? How to calculate $$4 \over {{x^4} + {y^4} + {z^4}}$$
from $$ x + y + z = 1, $$
$$ x^2 + y^2 + z^2 = 9, $$
$$ x^3 + y^3 + z^3 = 1. $$
Alternative answers:  A) $1 \over {33}$, B) $2 \over {33}$, C) $4 \over {33}$, D) $16 \over {33}$, E) $64 \over {33}.$

I tried to expanding $(x + y + z)^4 $ or ${(x^2 + y^2 + z^2)^2}$ and so then isolate $x^4 + y^4 + z^4$ but get too long expressions.
What notable product or special product could be applied to this problem?
 [/INST] \begin{align}
x^2+y^2+z^2&=(x+y+z)^2-2xy-2yz-2zx\\
9&=1-2(xy+yz+zx)\\
xy+yz+zx&=-4\\
(xy+yz+zx)^2&=16\\
x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)&=16\\
x^2y^2+y^2z^2+z^2x^2+2xyz&=16\qquad\text{since }x+y+z=1\\
\frac{(x^2+y^2+z^2)^2-(x^4+y^4+z^4)}{2}+2xyz&=16\tag{1}
\end{align}
On the other hand, from the identity
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
we have
\begin{align}
1-3xyz&=(1)\left(9-(-4)\right)\\
xyz&=-4
\end{align}
plugging it into $(1)$ we get
\begin{align}
\frac{9^2-(x^4+y^4+z^4)}{2}+2(-4)&=16\\
81-(x^4+y^4+z^4)-16&=32\\
x^4+y^4+z^4&=33
\end{align}
Then
$$\frac{4}{x^4+y^4+z^4}=\boxed{\color{blue}{\frac{4}{33}}}$$
</s>