Question

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

Answer 1

`(3y+1)^3+8y^3 = (5y+1)(7y^2+4y+1)`

`color(white)((3y+1)^3+8y^3) = 1/7(5y+1)(7y+2-sqrt(3)i)(7y+2+sqrt(3)i)`

Explanation:

The sum of cubes identity can be written:

`a^3+b^3=(a+b)(a^2-ab+b^2)`

Since both `(3y+1)^3` and `8y^3 = (2y)^3` are perfect cubes, we can use the sum of cubes identity with `a=(3y+1)` and `b=2y` as follows:

`(3y+1)^3+8y^3 = (3y+1)^3+(2y)^3`

`color(white)((3y+1)^3+8y^3) = ((3y+1)+2y)((3y+1)^2-(3y+1)(2y)+(2y)^2)`

`color(white)((3y+1)^3+8y^3) = (5y+1)((9y^2+6y+1)-(6y^2+2y)+4y^2)`

`color(white)((3y+1)^3+8y^3) = (5y+1)(7y^2+4y+1)`

The remaining quadratic factor has no linear factors with real coefficients.

`color(white)()`
Complex factors

If we do want to factor `7y^2+4x+1` then we can do it by completing the square and using the difference of squares identity, with non-real complex coefficients.

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

We can use this with `a=(7y+2)` and `b=sqrt(3)i` as follows:

`7(7y^2+4y+1) = 49y^2+28y+7`

`color(white)(7(7y^2+4y+1)) = (7y)^2+2(7y)(2)+2^2+3`

`color(white)(7(7y^2+4y+1)) = (7y+2)^2-(sqrt(3)i)^2`

`color(white)(7(7y^2+4y+1)) = ((7y+2)-sqrt(3)i)((7y+2)+sqrt(3)i)`

`color(white)(7(7y^2+4y+1)) = (7y+2-sqrt(3)i)(7y+2+sqrt(3)i)`

Dividing both ends by `7` we get:

`7y^2+4y+1 = 1/7(7y+2-sqrt(3)i)(7y+2+sqrt(3)i)`