# Solve for x

• Nov 19th 2009, 05:38 AM
BabyMilo
Solve for x
$\sqrt[3]{x^3+98}-x=2$

x should = 3 or -5.

thanks!
• Nov 19th 2009, 05:50 AM
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.
• Nov 19th 2009, 05:56 AM
BabyMilo
Quote:

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.

Quote:

Move -x to the rhs.
$\sqrt[3]{x^3+98}=2+x$

Quote:

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.
• Nov 19th 2009, 06:01 AM
Stuck Man
No, cube all of the rhs together. You have effectively done that to the lhs.
• Nov 19th 2009, 06:04 AM
BabyMilo
Quote:

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$
• Nov 19th 2009, 06:09 AM
BabyMilo
Worked it out thanks. common mistake. THanks
• Nov 19th 2009, 06:11 AM
Stuck Man
Yes. You can't break up the value and then raise the parts to a power.