Solve for x

**BabyMilo** · November 19th 2009, 04:38 AM

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

x should = 3 or -5.

thanks!

**Stuck Man** · November 19th 2009, 04:50 AM

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.

**BabyMilo** · November 19th 2009, 04:56 AM

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.

**Stuck Man** · November 19th 2009, 05:01 AM

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

**BabyMilo** · November 19th 2009, 05:04 AM

Originally Posted by Stuck Man

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

you mean this?
$<br /> (2+x)^3$

**BabyMilo** · November 19th 2009, 05:09 AM

Worked it out thanks. common mistake. THanks

**Stuck Man** · November 19th 2009, 05:11 AM

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

Thread: Solve for x