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

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(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 )...Show more

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

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...Show more

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!!...Show more

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...Show more