<s>[INST] Computing the unit normal vector - Simplifying help I have a surface
$$X(u,v) = \left(3uv^2 - u^3 - \frac{u}{3}, 3u^2v - v^3 - \frac{v}{3}, 2uv \right), $$
and the cross product
$$(X_u \times X_v) = \left(3(u^2 + v^2) \frac{1}{3} \right) \cdot \left(2v, 2u, \frac{1}{3} - 3(u^2 + v^2) \right).$$
I have to show that the unit normal vector is
$$N(u,v) = \frac{1}{3(u^2 + v^2) + \frac{1}{3}} \cdot \left( 2u, 2v, \frac{1}{3} - 3(u^2 + v^2) \right).$$
So, to find the unit, I just need to divide by $| X_u \times X_v|$ which should end up being $\left( 3(u^2 + v^2) + \frac{1}{3} \right)^2$. So I do this:
$$\sqrt{4u^2 + 4v^2 + \left(\frac{1}{3} - 3(u^2 + v^2) \right)^2}$$
$$ = \sqrt{4u^2 + 4v^2 + \frac{1}{9} -2(u^2 + v^2) + 9(u^2 + v^2)^2}$$
$$ = \sqrt{2u^2 + 2v^2 + \frac{1}{9} + 9(u^2 + v^2)^2}$$
but I'm stuck on how to simplify this. Expanding the $9(u^2 + v^2)^2$ doesn't really help me much. What should I do?
 [/INST] Expanding $(3(u^{2} + v^{2}) + \frac{1}{3})^{2}$, we see: 
$(3(u^{2} + v^{2}) + \frac{1}{3})*(3(u^{2} + v^{2}) + \frac{1}{3}) = 9(u^{2}+v^{2})^{2} + 2*3*\frac{1}{3}(u^{2} + v^{2}) + \frac{1}{9}$;
Does this help?
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