x^3 + 27y^6 = (x+b)(x^2-bx+b^2) then b=????

Answer 1

The second half is a sum of cubes. This is pre calculus stuff right. I might be able to help if there was any more info.

Does it say to factor or something else or does it just say solve?

Answer 2

Expand the right-hand side (using FOIL, say)

(x+b) (x^2 - bx + b^2) =
x (x^2 - bx + b^2) + b(x^2 - bx + b^2) =
(x^3 -bx^2 + b^2x) + (bx^2 -b^2x +b^3) =
x^3 + b^3

So this is the same as x^3 + 27y^6 = x^3 + b^3, which means that

b^3 = 27y^6.

Take the cube root of both sides -- the cube root of a product is the product of the cube roots, so

b = 3 y^2

Answer 3

x^3+27y^6 = (x+b)(x^2-bx+b^2)
=x^3+bx^2-bx^2-b^2x+b^2x+b^3
(you cancel out bx^2-bx^2, b^2x+b^2x) you get:

x^3+27y^6 = x^3+b^3, you cancel x^3
27y^6 = b^3
b^3 = 27y^6
b= square root of 27y^2
then, b=square root of 27y^6

Answer 4

b = 3y^2

you know that because the final term is b*b^2, which is 27y^6
b^3 = 27y^6
b = 3y^2