Question

How do you factor `8y^3+ 1`?

Answer 1

`(2y+1)(4y^2-2y+1)`

Explanation:

Given, `8y^3+1`
`rArr (2y)^3+(1)^3`
`rArr (2y+1)^3-3.2y.1(2y+1)`
`rArr (2y+1)[(2y+1)^2-6y]`
`rArr (2y+1)[4y^2+4y+1-6y]`
`rArr (2y+1)(4y^2-2y+1)`