# How do you factor y^3-3y^2+3y-1?

Sep 23, 2015

${y}^{3} - 3 {y}^{2} + 3 y - 1 = {\left(y - 1\right)}^{3}$

#### Explanation:

The 1, 3, 3, 1 pattern is a giveaway.

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

More generally, you can expand ${\left(a + b\right)}^{n}$ or recognise an expanded power of ${\left(a + b\right)}^{n}$ by looking at the appropriate row of Pascal's triangle: For example:

${\left(a + b\right)}^{5} = {a}^{5} + 5 {a}^{4} {b}^{3} + 10 {a}^{3} {b}^{2} + 10 {a}^{2} {b}^{3} + 5 a {b}^{4} + {b}^{5}$

Sep 23, 2015

${y}^{3} - 3 {y}^{2} + 3 y - 1 = {\left(y - 1\right)}^{3}$

#### Explanation:

A quick inspection reveals that if $y = 1$ then
$\textcolor{w h i t e}{\text{XXX}} {y}^{3} - 3 {y}^{2} + 3 y - 1 = 0$

Therefore $\left(y - 1\right)$ is a factor of ${y}^{3} - 3 {y}^{2} + 3 y - 1$

$\left({y}^{3} - 3 {y}^{2} + 3 y - 1\right) \div \left(y - 1\right) = {y}^{2} - 2 y + 1$
and
${y}^{2} - 2 y + 1 = \left(y - 1\right) \left(y - 1\right)$

Therefore
${y}^{3} - 3 {y}^{2} + 3 y - 1$
$\textcolor{w h i t e}{\text{XXX}} = \left(y - 1\right) \left(y - 1\right) \left(y - 1\right)$
$\textcolor{w h i t e}{\text{XXX}} = {\left(y - 1\right)}^{3}$

Note that this could also have been solved if you recognized the pattern of the given expression as a row from Pascal's triangle, but I thought a more general method of solution might be useful.