Factor using the formula for the sum or difference of two cubes: 125x^3 - 64.

Step-by-step solution

1. 1

   Confirm both terms really are cubes. $125=5^{3}$ and $x^{3}$ is already a cube, so $125x^{3}=(5x)^{3}$; and $64=4^{3}$. Since the operation between them is subtraction, the difference-of-cubes formula applies.

2. 2

   Write down the identity. $a^{3}-b^{3}=(a-b)\left(a^{2}+ab+b^{2}\right),$ with $a=5x$ and $b=4$. Note the middle sign in the quadratic factor is $+$, the opposite of the sign in the binomial factor.

3. 3

   Build the linear factor. $a-b=5x-4 .$

4. 4

   Build the quadratic factor. $a^{2}=(5x)^{2}=25x^{2},\qquad ab=(5x)(4)=20x,\qquad b^{2}=4^{2}=16,$ so the second factor is $25x^{2}+20x+16$. Its discriminant is $400-1600=-1200<0$, so it does not factor further over the real numbers.

5. 5

   Write the result and verify. $125x^{3}-64=(5x-4)\left(25x^{2}+20x+16\right).$ Expanding: $5x\cdot 25x^{2}=125x^{3}$, and every other term cancels in pairs, leaving $-4\cdot 16=-64$.