algebra 2, algebra 2, algebra 2????

(x-y^5)^3

= (x-y^5) x (x-y^5) x (x-y^5)

= (x-y x y x y x y x y ) x (x-y x y x y x y x y ) x (x-y x y x y x y x y )

(x - y^5)^3

= x^3 - 3x^2y^5 + 3xy^10 - y^15

For expanding cubic binomials the general formula is as follows:

(a + b) ^ 3 = a^3 + 3*a^2*b^1 + 3*a^1*b^2 + b^3

In your case, a is x and b is -y^5

So

(x - y^5)^3 = x^3 + 3*x^2*(-y^5)^1 + 3*x^1*(-y^5)^2 + (-y^5)^3

Simplified:

=x^3 - 3x^2*y^5 + 3x*y^10 - y^15

:D

Take your time, multiply each term and then combine like terms.

(x - y^5)^3 = (x - y^5)(x - y^5)(x - y^5)

It's easier to multiply only two expressions at a time.

(x - y^5)(x^2 - 2xy^5 + y^10)

x^3 - 2x^2 y^5 + xy^10 - x^2 y^5 + 2xy^10 - y^15

Combining terms

x^3 - 3x^2 y^5 + 3xy^10 - y^15

WOW!!

Since it is a quantity to the third power, multiply it by itself three times. (x - y^5)(x - y^5)(x - y^5) so you get (x^2 - 2xy^5 + y^10)(x - y^5) which then expands to:

x^3 - 2x^2y^5 + xy^10 - x^2y^5 + 2xy^10 - y^15

since

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

replace y with y^5 and you ge the answer.

(x – y^5)^3

=x3 – y15

(x-y^5)(x-y^5)(x-y^5)

=(x^2-2xy^5+y^10)(x-y^5)

=(x^3-2x^2y^5+xy^10-x^2-y^5+2xy^10-y^15)

=x^3-3x^2y^5+3xy^10-y^15