# How do I use the quadratic formula to solve 125x^3 - 1 = 0?

Jul 25, 2014

Hopefully you know what "difference of cubes" is, since this problem involves this method before you can use the quadratic formula... You'll see what I mean.

To begin, because you are subtracting you know that the difference of cubes should be in the form $\left({x}^{3} - {y}^{3}\right) = \left(x - y\right) \left({x}^{2} + x y + {y}^{2}\right)$.

To make this problem simpler, get your given equation in the form $\left({x}^{3} - {y}^{3}\right)$ . You should know that ${5}^{3} i s 125$ and ${1}^{3}$ is $1$ ... so rewrite the problem as follows:

${\left(5 x\right)}^{3} - {\left(1\right)}^{3}$

Now, you know that (5x) is the x value of the difference of cubes formula, and that 1 is the y value. So, sub these values into the formula (x - y)(x^2 + xy +y^2) like so:

$= \left(5 x - 1\right) \left[{\left(5 x\right)}^{2} + \left(5 x\right) \left(1\right) + {\left(1\right)}^{2}\right]$

Simplify it all:

$\left(5 x - 1\right) \left(25 {x}^{2} + 5 x + 1\right)$

Now, because the bracket on the left is in linear form (just x to the power of 1), you can solve it:

5x - 1 = 0
5x = 1
x = 1/5

That's your first solution. Now, you use the quadratic formula to solve the other bracket, since it is quadratic.