Expand (2a^2-3b^3)^2?

(2a^2-3b^3)^2=
(2a^2-3b^3) * (2a^2-3b^3) =
4a^4 -6a^2b^3 -6a^2b^3 + 9b^6 =
4a^4 -12a^2b^3 + 9b^6

You just have to foil. Since we're dealing with two sets of the same term, just make sure that both terms are multiplied by each other. Again, since we're dealing with two sets of the same term, we can use a trick:

square the first term, double the product of both terms together and square the last term. Or, if you want to look at it algebraically:

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

If you'll notice, we are dealing with that exact form, just with different coefficients and exponents. Not a problem though, just square 2a^2 to become 4a^4, double the product of 2a^2 and -3b^3, and so on. Because it is a difference of terms opposed to a summation, just remember the double product is the only term that will be negative. The first term is positive and the two negatives multiplied by each other become positive, so the last term is positive as well.

(2a^2 - 3b^3)^2
= (2a^2 - 3b^3)(2a^2 - 3b^3)
= 2a^2*2a^2 - 3b^3*2a^2 - 2a^2*3b^3 - 3b^3*3b^3
= 4a^4 - 6a^2b^3 - 6a^2b^3 - 9b^6
= 4a^4 - 12a^2b^3 - 9b^6

(2a² - 3b³) (2a² - 3b³)
4a^4 - 6a²b³ - 6a²b³ + 9b^6
4a^4 - 12a²b³ + 9b^6

Whenever you have (x + y)^2 it is always
(x + y)^2 = x^2 + 2xy + y^2

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

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

= 4a^4 - 6a^2b^3 + 9b^6