so.. hmm using the binomial theorem
[tex]\bf (x-3y)^8\implies
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(x)^8(-3y)^0\\
2&+8&(x)^7(-3y)^1\\
3&+28&(x)^6(-3y)^2\\
4&+56&(x)^5(-3y)^3\\
5&+70&(x)^4(-3y)^4
\end{array}[/tex]
now... hmmm notice... the first term in the binomial, starts off with a highest exponent of 8, in this case, and the exponent gradually goes down by 1 in each term
whilst for the second term in the binomial, is the opposite, starts off with an exponent of 0, and the exponent gradually goes up in each element
to get the coefficients for the expansion.... well, notice, the coefficient for the expanded 2nd element is always the exponent of the binomial, in this case 8
now, the next element's coefficient is, "the current coefficient, times the exponent of the first term, divided by the exponent of the second term in the next element"
to make it less muddy.... hmmm how did we get +28 for the 3rd element?
well, 8*7/2, coefficient * (first term's exponent) / (second term's exponent on following element)
how did we get 70 for the 5th term?
well 56*5/4
[tex]\bf 70(x)^4(-3y)^4\implies 70x^4[(-3)^4y^4]\implies 70x^4[81y^4]\implies +5670x^4y^4[/tex]
