Question 715610
There is a general formula that you were probably given, but probably makes no sense to you.
If your teacher wants it, write it the way it was given to you, but I will tell you what ideas I would use to get the answer (without worrying about formulas).
 
Since the power is {{{8}}} , there will be {{{8+1=9}}} terms.
I could start with {{{2^8=256}}} as a first term,
and end with {{{(-3y)^8}}} as a last term,
and that is probably what's expected.
 
All the terms will have combinations of powers of {{{2}}}
and powers of {{{(-3y)}}} with exponents adding up to {{{8}}}.
The exponents of {{{(-3y)}}} increase from term to term, starting at zero and going up to {{{8}}}.
The exponents of {{{2}}} decrease from {{{8}}} to zero.
The in-between terms, between the first term and the last term,
will also have combinatorial numbers as extra factors.
(There are many ways to write the symbols for combinatorial numbers,
but I do not know what way you would use in your class,
and I can only write combinatorial numbers in the form {{{(matrix(2,1,n,k))}}} on this website).
Those combinatorials are all combinations of {{{8}}} 
and the other number in the combination is one of the exponents.
Traditionally, it would be the exponent on {{{(-3y)}}}.
(If you used the exponent of {{{2}}} instead,
the symbol would look different,
but it would calculate as the same number).
 
So term number {{{4}}} would have
{{{(-3y)^((4-1))=(-3y)^3=-27y^3}}}
with {{{3}}} as the exponent for {{{(-3y)}}}.
It would also have
{{{2^((8-3))=2^5=32}}} and
it would also have the combinatorial number
{{{(matrix(2,1,8,3))=8*7*6/1/2/3= 56}}}
So the fourth term is
{{{56*32*(-27y^3)=-48384y^3}}}
 
The third term has {{{(matrix(2,1,8,2))=8*7/1/2=28}}}
{{{(-3y)^((3-1)))=(-3y)^2=9y^2}}} and {{{2^((8-2)))=2^6=64}}}
So the third term is {{{28*64*(9y^2)=16128y^2}}}
 
The second term has {{{(matrix(2,1,8,1))=8/1=8}}}
{{{(-3y)^((2-1)))=(-3y)^1=-3y}}} and {{{2^((8-1)))=2^7=128}}}
So the secondd term is {{{8*128*(-3y)=-3072y}}}
 
Then the first four terms would be
{{{256-3072y+16128y^2-48384y^3}}}
If you have to "show your work", you could just write
{{{2^8+8*2^7*(-3y)+(matrix(2,1,8,2))*2^6*(-3y)^2+(matrix(2,1,8,3))*2^5*(-3y)^3=256+8*128*(-3y)+28*64*(9y^2)+56*32*(-27y^3)=256-3072y+16128y^2-48384y^3}}}
 
A general formula for term number {{{k}}} of the expansion of {{{(a+b)^n}}}
is {{{(matrix(2,1,n,k))*a^((n-k+1))*b^((k-1))}}}