<s>[INST] If $a^2 + b^2 +c^2 =4$ and $a^3 + b^3 + c^3 = 8$ then find $a^4+b^4+c^4$ This is a popular question, but I can find $a^4+b^4+c^4$ if I know $a+b+c, a^2+b^2+c^2, a^3 +b^3 +c^3$ or in special case $a+b+c=0$ we only need a more condition like $a+b+c$. I have tried in many times and can't find out the way to archive it.
 [/INST] A somewhat simpler solution is given by Lagrange multipliers. The stationary points of $f(a,b,c)=a^3+b^3+c^3$ over $a^2+b^2+c^2=4$ occur where $3(a^2,b^2,c^2)=2\lambda(a,b,c)$, i.e. at $\pm\left(\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}\right)$ and at the permutations of $\pm(0,0,2)$ and $\pm(0,\sqrt{2},\sqrt{2})$. In particular
$$ \max_{a^2+b^2+c^2}a^3+b^3+c^3 = 8 $$
and in a point of absolute maximum we have $a^4+b^4+c^4=16$.
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