Answer
Maximum:$1$ and minimum: $\dfrac{1}{3}$
or,
Maximum:$f(0,0, \pm 1)= 1$ and minimum: $f(\pm \sqrt {\dfrac{1}{3}},\pm \sqrt {\dfrac{1}{3}},\pm \sqrt {\dfrac{1}{3}})=\dfrac{1}{3}$
Work Step by Step
Use Lagrange Multipliers Method:
$\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$
This yields $\nabla f=\lt 4x^3,4y^3,4z^3 \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y,2z\gt$
Using the constraint condition $x^2+y^2+z^2=1$ we get, $4x^3=\lambda 2x ,4y^3=\lambda 2y,4z^3=\lambda 2z$
After solving, we get $x^2=y^2=z^2$
Since, $g(x,y)=x^2+y^2+z^2=1$ gives $x=\pm \dfrac{1}{\sqrt 3}$
Thus, $x=y=z=\pm \sqrt {\dfrac{1}{3}}$ .This yields eight different points, but each one has a minimum value $\dfrac{1}{3}$
Also, we have possible points: $x=y=0,z=\pm 1$. This yields two different points, but each one has a maximum value of $1$
Hence, maximum:$1$ and minimum: $\dfrac{1}{3}$
or,
Maximum:$f(0,0, \pm 1)= 1$ and minimum: $f(\pm \sqrt {\dfrac{1}{3}},\pm \sqrt {\dfrac{1}{3}},\pm \sqrt {\dfrac{1}{3}})=\dfrac{1}{3}$
