# How do you simplify  (y-4)^3?

May 28, 2017

${\left(y - 4\right)}^{3}$ is simplified.

Expanded: ${\left(y - 4\right)}^{3} = {y}^{3} - 12 {y}^{2} + 48 y - 64$

#### Explanation:

${\left(y - 4\right)}^{3}$ is simplified.

However, if you want to expand it using binomial expansion, you can use Pascal's triangle:

Pascal's triangle gives the combinations:

$\text{ } 1$ $\text{ } {\left(a + b\right)}^{0}$
$\text{ "1 " " 1}$ $\text{ } {\left(a + b\right)}^{1}$
$\text{ "1 " "2" " 1}$ $\text{ } {\left(a + b\right)}^{2}$
$\text{ "1 " " 3 " " 3 " } 1$ $\text{ } {\left(a + b\right)}^{3}$

Binomial Theorem: ${\left(a + b\right)}^{n} =$

 ""_nC_0 a^n b^0 + _nC_1 a^(n-1) b^1 + _nC_2a^(n-2)b^2 + ... + _nCn a^0 b^n

where  ""_nC_0 = (n!)/((n-0!)(0!)) = 1; " " _nC_1 = (n!)/((n-1!)(1!))

 ""_nC_2 = (n!)/((n-2!)(2!)); " "_nC_n = (n!)/((n-n!)(n!)) = 1

Given: simplify ${\left(y - 4\right)}^{3}$

Let $a = y \text{ and " b = -4 ; " } n = 3$

From Pascal's triangle: ${\text{_3C_0 = ""_3C_3 = 1; ""_3C_1 = }}_{3} {C}_{2} = 3$

${\left(y - 4\right)}^{3} =$
${\text{_3C_0 y^3 (-4)^0 + ""_3C_1 y^2 (-4)^1 + ""_3C_2 y^1 (-4)^2 + }}_{3} {C}_{3} {y}^{0} {\left(- 4\right)}^{3} =$

${y}^{3} + 3 {y}^{2} \left(- 4\right) + 3 y \left(16\right) - 64 =$

${y}^{3} - 12 {y}^{2} + 48 y - 64$