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

Jun 13, 2018

$\textcolor{b l u e}{{\left(6 {y}^{4} - 1\right)}^{2} = \left(36 \cdot {y}^{8} - 12 {y}^{4} + 1\right)}$

#### Explanation:

Note that:

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

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

now lets expand ${\left(6 {y}^{4} - 1\right)}^{2}$

$\textcolor{b l u e}{{\left(6 {y}^{4} - 1\right)}^{2} = \left(36 \cdot {y}^{8} - 12 {y}^{4} + 1\right)}$

Jun 13, 2018

${\left(6 {y}^{4} - 1\right)}^{2} = 36 {y}^{8} - 12 {y}^{4} + 1$

#### Explanation:

The square of a sum can be expanded as

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

Which means that you have to square both terms, and add twice their product.

The two terms are $6 {y}^{4}$ and $- 1$. Their squares are $36 {y}^{8}$ and $1$.

Their product is $- 6 {y}^{4}$, so twice their product is $- 12 {y}^{4}$

Now we only need to sum everything together to get

${\left(6 {y}^{4} - 1\right)}^{2} = 36 {y}^{8} - 12 {y}^{4} + 1$