Find coefficients with the binomial theorem I need the find the coefficient of $$x^6y^6 \,\ \text{in} \,\ (2x^3-3y^2)^5$$
I don't know how to do these style of problems, where there are powers within the parenthesis. I know how to do the ones where there's just a power outside of the parenthesis, but I am lost here, seems like the technique to complete these are different.  
 A: Hint: In the binomial expansion formula:
$$(a + b)^5 = \sum_{i = 0}^5 \binom{5}{i}a^{5 - i}b^i,$$
the only possibility you get $x^6y^6$ term is when $i = 3$ (why?), now find $a$ and $b$ under your case and compute the coefficient.
A: $$(2x^3-3y^2)^5=\sum_{k=0}^5 (^5_k)(2x^3)^k(-3y^2)^{5-k}$$
$$=\sum_{k=0}^5 (^5_k)2^k (-3)^{5-k} x^{3k} y^{10-2k}$$
Now $3k=6$ and $10-2k=6$ implies $k=2$
So the answer is $(^5_2)2^2 (-3)^{5-2}=10 \cdot 4 \cdot (-27) =-1080$
A: From the binomial theorem you have
$$(2x^3-3y^2)^5=\sum_{k=0}^5\binom5k(2x^3)^k(-3y^2)^{5-k}=\sum_{k=0}^5\binom5k2^k(-3)^{5-k}x^{3k}y^{10-2k}\;,$$
so you want the coefficient $\dbinom5k2^k(-3)^{5-k}$ when $3k=6$ and $10-2k=6$. If there is such a $k$, use it to compute the coefficient; if there isn’t, then there is no $x^6y^6$ term, and the coefficient is therefore $0$.
A: The binomial theorem is:
$(a + b)^n$ = $\sum_{i=0}^n {n \choose i}a^ib^{n-i} = \sum_{i=0}^n \frac{n!}{i!(n-i)!} a^ib^{n-i}$
Which for n = 5 is
$a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5$
So for $(2x^3 - 3y^2)^5$ you want to find ${5 \choose i}2^ix^{3i}(-3)^{5-i}y^{2(5-i)}$ where $3i = 2(5-i) = 6$.  That is, $i = 2$.
So you want  ${5 \choose 2}2^2x^{6}(-3)^{3}y^{6} = 10*4*(-27)x^6y^6$.  The coefficient is $10* 4(-27) =  -1080 $.
Sheesh!
