Finding the maximum value If $a^2+b^2+c^2+d^2=4$ are real numbers, then how to find the maximum value for:$$a^3+b^3+c^3+d^3$$

To maximize/minimize $a^3+b^3+c^3+d^3$ under the constraint that $a^2+b^2+c^2+d^2=4$ means we need
$$
a^2\delta a+b^2\delta b+c^2\delta c+d^2\delta d=0\tag{1}
$$
for any deltas so that
$$
a\,\delta a+b\,\delta b+c\,\delta c+d\,\delta d=0\tag{2}
$$
Standard orthogonality arguments say that there must be a $\lambda$ so that
$$
(a^2,b^2,c^2,d^2)=\lambda\,(a,b,c,d)\tag{3}
$$
That is, $(a^2,b^2,c^2,d^2)$ must be perpendicular to any delta which is perpendicular to $(a,b,c,d)$.
$(3)$ says that $a^2-\lambda a=0$. Thus, either $a=0$ or $a=\lambda$. The same is true for $b$, $c$, and $d$.
There are a finite number of possibilities, each depending on how many of $b$, $c$, and $d$ are zero and whether $\lambda$ is positive or negative.
$0$ zeroes: $\lambda=\pm1$ and $a^3+b^3+c^3+d^3=\pm4$
$1$ zero: $\lambda=\pm\sqrt{\frac43}$ and $a^3+b^3+c^3+d^3=\pm4\sqrt{\frac43}$
$2$ zeroes: $\lambda=\pm\sqrt{2}$ and $a^3+b^3+c^3+d^3=\pm4\sqrt{2}$
$3$ zeroes: $\lambda=\pm2$ and $a^3+b^3+c^3+d^3=\pm8$
Thus, the maximum of $a^3+b^3+c^3+d^3$ is $8$.
