# Math Help - Solve for x

1. ## Solve for x

$\sqrt[3]{x^3+98}-x=2$

x should = 3 or -5.

thanks!

2. Move -x to the rhs. Cube the rhs and remove the cube root on the lhs. You will probably know how to solve it from there.

3. Originally Posted by Stuck Man
Move -x to the rhs. Cube the rhs and remove the cube root on the lhs. You will probably know how to solve it from there.
Move -x to the rhs.
$\sqrt[3]{x^3+98}=2+x$

Cube the rhs and remove the cube root on the lhs.
${x^3+98}=2^3+x^3$

do you mean this?

this cant be right as the $x^3$ cancel themselves.

4. No, cube all of the rhs together. You have effectively done that to the lhs.

5. Originally Posted by Stuck Man
No, cube all of the rhs together. You have effectively done that to the lhs.
you mean this?
$
(2+x)^3$

6. Worked it out thanks. common mistake. THanks

7. Yes. You can't break up the value and then raise the parts to a power.