I have this easy question please:
Find the maximum and minimum values of the function $ (, , ) = ^3 + ^3 + ^3$ on the plane $ + + = 4$ by Lagrange multipliers method.
My attempt is shown in the picture. I reached to nine possiblitiest from which I need to substitute for $\lambda$ to find the min/max. But when I find $\lambda$, I got two posiiblilities with two values of $ (, , )$.



No, it is not correct.
You have the system$$\left\{\begin{array}{l}3x^2=\lambda\\3y^2=\lambda\\3z^2=\lambda\\x+y+z-4=0.\end{array}\right.$$So, for some number $\lambda$ (which has to be greater than or equal to $0$), you have that each of the numbers $x$, $y$, and $z$ is equal to $\pm\sqrt{\frac\lambda3}$. Since they're all equal and their sum is greater than $0$, then either all of them are equal to $\sqrt\frac\lambda3$ or two of them are equal to $\sqrt\frac\lambda3$, whereas the third one is equal to $-\sqrt\frac\lambda3$.
If they are all equal to $\sqrt\frac\lambda3$, then$$4=x+y+z=3\sqrt\frac\lambda3=\sqrt{3\lambda},$$and therefore $\lambda=\frac{16}3$ and $x=y=z=\frac43$. Otherwise$$4=x+y+z=\sqrt{\frac\lambda3},$$and therefore $\lambda=48$ and$$(x,y,z)=(-4,4,4),\ (x,y,z)=(4,-4,4)\text{, or }(x,y,z)=(4,4,-4).$$Can you take it from here?