3x^1/3 -16x^1/3 -35=0

I substituted x^1/3 for u and end up getting stuck at

3u^2 -16u-35 What do I do from here?

Originally Posted by daword177
3x^2/3 -16x^1/3 -35=0
I substituted x^1/3 for u and end up getting stuck at
3u^2 -16u-35 What do I do from here?
$3u^2-16u-35=(3u+5)(u-7)$

That's a quadratic equation (once you add the "= 0") so if you cannot factor, complete the square or use the quadratic formula.

Originally Posted by daword177
3x^1/3 -16x^1/3 -35=0

I substituted x^1/3 for u and end up getting stuck at

3u^2 -16u-35 What do I do from here?
Do you mean

3x^2/3 -16x^1/3 -35=0

If so: Your substitution is OK.

Use the quadratice formula to solve the equation

$3u^2 -16u-35 = 0$

for u.

Afterwards plug in the values of u into the equation

$\sqrt[3]{x} = u$

and solve for x.