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

##### 1 Answer

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

To make this problem simpler, get your given equation in the form ** #(x^3 - y^3)#** . You should know that

**and**#5^3 is 125#

**is**#1^3#

**... so rewrite the problem as follows:**#1#

#(5x)^3 - (1)^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:

Simplify it all:

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.

Recall the quadratic formula is:

Now sub in all your values from 25x^2 + 5x +1. **a = 25** , **b = 5** , and **c = 1** .

Now solve: **WAIT!** If you **check the discriminant** (under the root sign), you'll notice that **it comes out to be negative** ! This means there are **no real roots**! If the discriminant was NOT NEGATIVE, you would solve for the roots as usual and you would have 3 roots!

As such, the one root you got before (x=1/5) is the only real root! Hopefully you understood all this and hopefully I was of some help!