Find the coefficient of $x^9$ in the expansion of $(1+9x+27x^2+27x^3)^6$

Question

Find the coefficient of $x^9$ in the expansion of $(1+9x+27x^2+27x^3)^6$.

---

Consider, $$(1+9x+27x^2+27x^3)^6 = \sum \frac{6!}{a!\cdot b!\cdot c!\cdot d!}(1^{a}\cdot9x^{b} \cdot (27x^2)^{c} \cdot (27^3)^d)$$

Power of $x=x^{b+2c+d}$

So,

• $b +2c +3d = 9$

Also,

• $a +b+c+d = 6$

Are we supposed to find different combination of $a,b,c $ and $d$ in such a way that the above $2$ equations are satisfied and then add the coefficients of the different combinations?

I don't know how to proceed further. Any Hint would be appreciated.

Answer 1

First note that $1+9x+27x^2+27x^3=(1+3x)^3$. So the problem is the same as calculating the coefficient of $x^9$ in $(1+3x)^{18}$. Which is much easier.