SOLUTION: Use binomial expansion to expand the following (2x-3y)^4

The first term is (2x) the second term is (-3y).
The power is 4 so write "(2x) (-3y)" one more than 4, that is, 5 times:

 (2x) (-3y)  +  (2x) (-3y)  +  (2x) (-3y)  +  (2x) (-3y)  +  (2x) (-3y) 

Give the first factor (2x) a 4 power and the next (2x) a 3 power and so
on down to the 0 power on the last term:

 (2x)4(-3y)  +  (2x)3(-3y)  +  (2x)2(-3y)  +  (2x)1(-3y)  +  (2x)E(-3y) 

Give the second factor (-3y) a 0 power and the next (-3y) a 1 power and so
on up to the 4 power on the last term:

 (2x)4(-3y)0 +  (2x)3(-3y)1 +  (2x)2(-3y)2 +  (2x)1(-3y)3 +  (2x)0(-3y)4

Give the first term the coefficient 1

1(2x)4(-3y)0 +  (2x)3(-3y)1 +  (2x)2(-3y)2 +  (2x)1(-3y)3 +  (2x)0(-3y)4

Multiply that coefficient 1 by the exponent of the first factor in the 1st
term, which is 4, getting 4. Then divide that by the number of term which those
came from, which is term number 1, so you divide 4 by 1 getting 4 and then write
that 4 as the coefficient of the second term:

1(2x)4(-3y)0 + 4(2x)3(-3y)1 +  (2x)2(-3y)2 +  (2x)1(-3y)3 +  (2x)0(-3y)4

Multiply that coefficient 4 by the exponent of the first factor in the 2nd 
term, which is 3, getting 12. Then divide that by the number of term which those
came from, which is term number 2, so you divide 12 by 2 getting 6 and then
write that 6 as the coefficient of the third term:

1(2x)4(-3y)0 + 4(2x)3(-3y)1 + 6(2x)2(-3y)2 +  (2x)1(-3y)3 +  (2x)0(-3y)4

Multiply that coefficient 6 by the exponent of the first factor in the 3rd 
term, which is 2, getting 12. Then divide that by the number of term which those
came from, which is term number 3, so you divide 12 by 3 getting 4 and then
write that 4 as the coefficient of the fourth term:

1(2x)4(-3y)0 + 4(2x)3(-3y)1 + 6(2x)2(-3y)2 + 4(2x)1(-3y)3 +  (2x)0(-3y)4

Multiply that coefficient 4 by the exponent of the first factor in the 4th 
term, which is 1, getting 4. Then divide that by the number of term which those
came from, which is term number 4, so you divide 4 by 4 getting 1 and then
write that 4 as the coefficient of the fifth, and last term:

1(2x)4(-3y)0 + 4(2x)3(-3y)1 + 6(2x)2(-3y)2 + 4(2x)1(-3y)3 + 1(2x)0(-3y)4

Now we simplify. We erase the 1 coefficients and 1 exponents 

 (2x)4(-3y)0 + 4(2x)3(-3y)  + 6(2x)2(-3y)2 + 4(2x)(-3y)3  +  (2x)0(-3y)4

Anything to the zero power is 1 so we erase the factors with 0 exponents

 (2x)4       + 4(2x)3(-3y)  + 6(2x)2(-3y)2 + 4(2x)(-3y)3  +       (-3y)4

We observe that negtive numbers raised to even powers are positive and
negative numbers raised to odd powers are negative and so rewrite the
signs as alternating: 

 (2x)4       - 4(2x)3(3y)   + 6(2x)2(3y)2  - 4(2x)(3y)3   +       (3y)4

  24x4       - 4*223x3*3y   +  6*22x2*32y2 -  4*2x*33y3   +        34y4

  16x4       - 4*8x3y       +  6*4x2*9y2   -   8x*27y3    +        81y4

  16x4       -  96x3y       +   216x2y2    -   216xy3     +        81y4

  16x4       -  96x3y       +   216x2y2    -   216xy3     +        81y4

Edwin