# 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.

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!

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.

• Everyone else says i=2... – Zion Todd Nov 4 '15 at 20:05
• If you observe their and my formulae closely, you will find they are slightly different. – Zhanxiong Nov 4 '15 at 20:06

$$(2x^3-3y^2)^5=\sum_{k=0}^5 (^5_k)(2x^3)^k(-3y^2)^{5-k}$$ $$=\sum_{k=0}^5 (^5_k)2^k (-3)^{5-k} x^{3k} y^{10-2k}$$

Now $3k=6$ and $10-2k=6$ implies $k=2$

So the answer is $(^5_2)2^2 (-3)^{5-2}=10 \cdot 4 \cdot (-27) =-1080$

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 $0$.