# Can someone solve this equation for a?

1. Apr 28, 2005

### Icebreaker

Can someone solve this equation for a?

$$b^2a+2bc^2=\frac{a^3b^2}{c^2}+2ba^2+c^2a$$

2. Apr 28, 2005

### AKG

P(x) = (b/c)²x³ + (2b)x² + (c² - b²)x + (-2bc²)

We want to find a such that P(a) = 0

Factor this cubic P, and you'll get something in the form P(x) = (x - a)(dx² + ex + f)

That a is precisely the one you're looking for. How to factor cubics.