# How do you expand (2x^4-1)^6?

Nov 24, 2016

To make it a bit easier to view:

Let $a = 2 {x}^{4} \mathmr{and} b = - 1$

Then we will have ${\left(a + b\right)}^{6}$

To start expanding, we should separate the function:

$= \left(a + b\right) \left(a + b\right) \left(a + b\right) \left(a + b\right) \left(a + b\right) \left(a + b\right)$

Then start multiplying:

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

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

Repeat this process until all values are distributed (no parentheses)

You should end up with:

$64 {x}^{24} - 192 {x}^{20} + 240 {x}^{16} - 160 {x}^{12} + 60 {x}^{8} - 12 {x}^{4} + 1$