factor y^3+27?

This is the sum of two cubes:

(y³ + 3³) = (y + 3)(y² - 3y + 9)

Since both terms are perfect cubes, the binomial can be factored using the sum of cubes formula:

(y+3)(y2-3y+9)

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http://mathway.com/problem.aspx?p=basicmath

x³+y³ = (x+y)(x²-xy+y²)

y³+3³ = (y+3)(y²-3y+9)

y²-3y+9 has no real roots.

(3±√9-36)/2 = 3/2 ± (√-27)/2 = (3±3i√3)/2

Factors: y=-3, (3+3i√3)/2, (3-3i√3)/2

use formula
a^3+b^3=(a+b)(a^2-ab+b^2)
y^3+27
=y^3+ 3^3
=(y+3)(y^2-3y+9) ans

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

y^3 + 27
= (y + 3)(y^2 - 3y + 9)

1) cube root each term, place in first brackets.
2)
a) square the first
b) negative product
c) square the second.

y^3 + 27

(y + 3) (? + ? + ?)

"Square the first" (square y)

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

"Negative product" (of 3 and y).
Take the product of 3 and y, and then negate the result by multiplying it by (-1).

(y + 3)(y^2 - 3y + ?)

"Square the second" (square 3)

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