If $ x^2+y^2+z^2 =1$ for $x,y,z \in \mathbb{R}$, then find maximum value of $ x^3+y^3+z^3-3xyz $. If $ x^2+y^2+z^2 =1$, for $x,y,z \in \mathbb{R}$, what is the maximum of 
$ x^3+y^3+z^3-3xyz $ ?
I factorize it... Then put the maximum values of $x+y+z$ and min value of $xy+yz+zx$...
But it is wrong as they don't hold simultaneously!
Also can it be solved using partial differentiation ?
And plz provide a solution without it.!

Yet another way, let $p = x+y+z, q = xy+yz + zx$.  Then $x^2+y^2+z^2 = p^2-2q=1$ and  $x^3+y^3+z^3-3xyz = p^3-3pq = \frac12p(3-p^2)$ .  Also $x^2+y^2+z^2 = 1 \implies p^2\le 3$.
So we want the maximum of $\frac12p(3-p^2)$, subject to $p^2 \le 3$, which easily$^\dagger$ gives $p=1$.  

$^\dagger$ Equivalently this is when $\frac14p^2(3-p^2)^2$ gets maximised, or $(p^2)\dfrac{3-p^2}2 \dfrac{3-p^2}2$.  As the last is a product of three positive terms with constant sum $3$, it gets maximised when all terms are equal, i.e. $p=1$.
