If B= A^3 / (A -2) what does 'A' equal?

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- Aug 8th 2011, 01:42 PMwolfA^3 / (A -2) = B Could someone solve this for 'A' ?
If B= A^3 / (A -2) what does 'A' equal?

- Aug 8th 2011, 01:54 PMSironRe: A^3 / (A -2) = B Could someone solve this for 'A' ?
Do you want an expression like: $\displaystyle A=...$ so in function of B?

- Aug 8th 2011, 02:11 PMwolfRe: A^3 / (A -2) = B Could someone solve this for 'A' ?
Yes, I would like the equation solved in terms of A.

For example, if 2A^2 = B, then A = square root (B/2) - Aug 8th 2011, 02:40 PMSironRe: A^3 / (A -2) = B Could someone solve this for 'A' ?
My first impression says it's not possible.

- Aug 8th 2011, 02:51 PMPlatoRe: A^3 / (A -2) = B Could someone solve this for 'A' ?
There are some conditions under which a cubic can be solved.

Here is a discussion. - Aug 8th 2011, 02:53 PMQuackyRe: A^3 / (A -2) = B Could someone solve this for 'A' ?
Possible, but not fun, according to my cheating. It depends on why you'd need it.

- Aug 8th 2011, 02:54 PMSironRe: A^3 / (A -2) = B Could someone solve this for 'A' ?
- Aug 8th 2011, 04:09 PMwolfRe: A^3 / (A -2) = B Could someone solve this for 'A' ?
Thank you Siron, Plato and especially Quacky for the link to the answer.

At first that seems as if it's an easy equation to solve but*not*after looking at that answer.

Thanks everyone. - Aug 9th 2011, 03:35 AMBacteriusRe: A^3 / (A -2) = B Could someone solve this for 'A' ?
Yes polynomials do tend to "look" easy to solve because they have nice round integers as coefficients... but in general solving a polynomial algebraically is not easy. For instance solving $\displaystyle x^5 - 1 = 0$ is easy, make that $\displaystyle x^5 + 2x - 1 = 0$ and we're all stumped. In fact in practice polynomials are solved by approximation in the real world. But that doesn't mean it's wrong to learn how to solve cubics!