Difference of Cubes

${x}^{3}-64$ has no pairs to group, but it has a pattern: ${x}^{3}$ is x cubed and 64 is 4 cubed. Any difference of two cubes factors the same way:

$${a}^{3}-{b}^{3} = \left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$$

With a = x and b = 4, that gives ${x}^{3}-64 = \left(x-4\right)\left({x}^{2}+4x+16\right)$. Multiply it out to see why it works: x times the long bracket is ${x}^{3}+4{x}^{2}+16x$, and -4 times it is $-4{x}^{2}-16x-64$. The middle terms cancel, leaving ${x}^{3}-64$.

The quadratic factor ${x}^{2}+4x+16$ has no real factors: its discriminant, 16 - 64, is negative, so its two roots are complex. That makes 4 the only real root of ${x}^{3}-64$. Learn the cubes 1, 8, 27, 64 and 125 so you can spot the pattern.