Question

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

Answer 1

`y^3-3y^2+3y-1 = (y-1)^3`

Explanation:

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

In general, `(a+b)^3 = a^3+3a^2b+3ab^2+b^3`

More generally, you can expand `(a+b)^n` or recognise an expanded power of `(a+b)^n` by looking at the appropriate row of Pascal's triangle:

For example:

`(a+b)^5 = a^5+5a^4b^3+10a^3b^2+10a^2b^3+5ab^4+b^5`

Answer 2

`y^3-3y^2+3y-1 = (y-1)^3`

Explanation:

A quick inspection reveals that if `y=1` then
`color(white)("XXX")y^3-3y^2+3y-1 = 0`

Therefore `(y-1)` is a factor of `y^3-3y^2+3y-1`

`(y^3-3y^2+3y-1)div(y-1) = y^2-2y+1`
and
`y^2-2y+1 = (y-1)(y-1)`

Therefore
`y^3-3y^2+3y-1`
`color(white)("XXX")=(y-1)(y-1)(y-1)`
`color(white)("XXX")=(y-1)^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.