The binomial theorem is as follows:

\((a+b)^2=\sum_{k=0}^{n}\binom{n}{k}a^{(n-k)}b^k \)

The formula never helped me much either, but here is the explanation. Looking at any expansion, there are coefficients and exponents. For example, the expansion for \((a+b)^3\) is as follows:

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

Just looking at the coefficients, you get 1, 3, 3, and 1. This is the fourth row of the Pascal triangle.

For the exponents, you get 3, 2, 1, 0 for a; and 0, 1, 2, 3 for b. From this, we can deduce what \((2x-3y)^5\) will be.

**The expansion for just **\((x-y)^5\):

The exponents for x will be 5, 4, 3, 2, 1, 0 and the exponents for y will be 0, 1, 2, 3, 4, 5.

The coefficients will be the sixth row of the Pascal triangle, which is 1, 5, 10, 10, 5, 1

Therefore, the expansion will be \((x-y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5\)

However, we are not done yet. We need to add the positive/negative signs, alternating.

\((x-y)^5=x^5-5x^4y+10x^3y^2-10x^2y^3+5xy^4-y^5\)

Using this equation, you can plug in the values to find the given expression.

I hope this helped,

Gavin.