# How do you write the expression (y-3)(y-3)(y-3) using exponents?

Jan 11, 2017

$\left(y - 3\right) \left(y - 3\right) \left(y - 3\right) = {\left(y - 3\right)}^{3} = {y}^{3} - 9 {y}^{2} + 27 y - 27$

#### Explanation:

This is a product of three binomials, who are all the same i.e. $\left(y - 3\right) \left(y - 3\right) \left(y - 3\right)$.

We can write it simply as ${\left(y - 3\right)}^{3}$.

For actual mutiplication, we use distributive property, first over $\left(y - 3\right) \left(y - 3\right)$ and then multiplying the result by $\left(y - 3\right)$

$\left(y - 3\right) \left(y - 3\right)$

= $y \left(y - 3\right) - 3 \left(y - 3\right)$

= $y \times y - 3 \times y - 3 \times y - 3 \times \left(- 3\right)$

= ${y}^{2} - 3 y - 3 y + 9$

= ${y}^{2} - 6 y + 9$

and now multiplying above again by $\left(y - 3\right)$

$\left(y - 3\right) \left(y - 3\right) \left(y - 3\right)$

= $\left(y - 3\right) \left({y}^{2} - 6 y + 9\right)$

= $y \left({y}^{2} - 6 y + 9\right) - 3 \left({y}^{2} - 6 y + 9\right)$

= $y \times {y}^{2} - 6 y \times y + 9 \times y - 3 \times {y}^{2} - 3 \times \left(- 6 y\right) - 3 \times 9$

= ${y}^{3} - 6 {y}^{2} + 9 y - 3 {y}^{2} + 18 y - 27$

= ${y}^{3} - 9 {y}^{2} + 27 y - 27$