$\displaystyle \sqrt[3]{x^3+98}-x=2$ x should = 3 or -5. thanks!
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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.
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. $\displaystyle \sqrt[3]{x^3+98}=2+x$ Cube the rhs and remove the cube root on the lhs. $\displaystyle {x^3+98}=2^3+x^3$ do you mean this? this cant be right as the $\displaystyle x^3$ cancel themselves.
No, cube all of the rhs together. You have effectively done that to the lhs.
Originally Posted by Stuck Man No, cube all of the rhs together. You have effectively done that to the lhs. you mean this? $\displaystyle (2+x)^3$
Worked it out thanks. common mistake. THanks
Yes. You can't break up the value and then raise the parts to a power.
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