factoring help?

Question

54x³ +2y³

Answer 1

Your first goal is to pull out the common factor of 2.

2(27x^3 + y^3)

Now, what you have is what's called a sum of cubes. Sums and differences of cubes have a factoring method that goes as follows.

(1) In one bracket, take the cube root of each term. (In the above case, you take the cube root of 27x^3, which is 3x, and the cube root of y^3, which is y). What you'd have is, in one bracket,

(3x + y)

(2). Remember these following steps for the second bracket, which is going to contain three terms. "square the first, negative product, square the last".

"Square the first" That means to take the first term in the first set of brackets, and square it. (3x)^2 = 9x^2.

(3x + y) (9x^2 + ? + ?)

"Negative product" This means to take the product of both terms in the first set of brackets, and then multiply that product by -1. In this case, (3x)(y) = 3xy, multiplied by -1 is -3xy

(3x + y) (9x^2 - 3xy + ?)

"Square the last". This means to square the second term in the first set of brackets. In this case, it would be y^2.

(3x + y) (9x^2 - 3xy + y^2)

Now, remember there was a lingering 2 out there, so your whole answer would be

2(3x + y) (9x^2 - 3x6 + y^2)

Answer 2

2(27x^3 + y^3) =

2(3x + y)(9x^2 - 3xy + y^2)

Answer 3

= 2(27 x3 + y3 )
= 2[(3x)3 + (y)3]
= 2[(3x+y)3 – 3(3x)(y)(3x+y)]
= 2[(3x+y)3 – 9xy(3x+y)]
= 2(3x+y)[(3x+y)^2 - 9xy)]
= 2(3x+y)(9x^2+ y^2 - 3xy)