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.!

Let $f(t)=\frac{t}{2}(3-t^2)$. It is straightforward to check that $f$ attains its maximum on the interval $[-\sqrt{3},\sqrt{3}]$ at $t=1$, with $f(1)=1$.
Now, if $x^2+y^2+z^2=1$ then
$$\eqalign{
x^3+y^3+z^3-3xyz&=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\cr
&=\frac{ x+y+z }{2}\left(3(x^2+y^2+z^2)-(x+y+z)^2\right)\cr
&=f(x+y+z)
}
$$
But by Cauchy-Schwarz inequality we have
$$|x+y+z|\leq\sqrt{3}\sqrt{x^2+y^2+z^2}=\sqrt{3}$$
So,
$$
x^3+y^3+z^3-3xyz\leq f(1)=1
$$
with equality if and only if $x+y+z=1$. This corresponds to the points of the circle
of intersection of the sphere of equation $x^2+y^2+z^2=1$ with the plane of equation
$x+y+z=1$.
