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

1 Answer
Aug 20, 2017

Answer:

#(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)#