The first thing we have to do is find the pascal coefficients of the triangle for this we use the following image
From our exercise we know that
[tex]\begin{gathered} (2x^3-3y)^3\to(a+b)^3 \\ a=2x^3 \\ b=-3y \end{gathered}[/tex]
Then our coefficients of Pascal's triangle are 1 - 3 - 3 - 1
Using the right triangle we get:
[tex]1a^3+3a^2b+3ab^2+1b^3[/tex]
We substitute the values of a and b in our new expression to find our solution
[tex]1(2x^3)^3+3(2x^3)^2(-3y)+3(2x^3)(-3y)^2+1(-3y)^3[/tex][tex]\begin{gathered} 8x^9+3(4x^6)^{}(-3y)+3(2x^3)(9y^2)^{}-27y^3 \\ 8x^9-36x^6y+54x^3y^2-27y^3 \end{gathered}[/tex]
So the solution is:
[tex]8x^9-36x^6y+54x^3y^2-27y^3[/tex]
