Consider the expansion of (3x^2 - \frac{1}{x})^9. Find the constant term in this expansion.

Question:

Consider the expansion of {eq}(3x^2 - \frac{1}{x})^9. {/eq} Find the constant term in this expansion.

Binomial Theorem:

The Binomial Theorem states the following formula for expanding a binomial {eq}(a+b) {/eq} raised to the power {eq}n {/eq}, where {eq}n {/eq} is a natural number:

{eq}\displaystyle (a+b)^n = \bigg(\sum_{k=0}^n \binom nka^kb^{n-k}\bigg) {/eq}

{eq}\displaystyle \binom nk = \frac {n!}{k!(n-k)!} {/eq} equals the number of ways of choosing {eq}k {/eq} objects in any order from {eq}n {/eq} objects without replacement.

Expanding the given expression using the binomial theorem, we obtain

{eq}\begin{align*} \left(3x^2 - \frac{1}{x}\right)^9 &= \sum_{k=0}^9 \frac {9!}{k!(9-k)!}(3x^2)^k\left(-\frac{1}{x}\right)^{9-k}\\[2ex] &=\sum_{k=0}^9 \frac {9!}{k!(9-k)!}3^k(-1)^{9-k}x^{2k-(9-k)}\\[2ex] &=\color{DarkOrange}{\sum_{k=0}^9 \frac {9!}{k!(9-k)!}3^k(-1)^{9-k}x^{3k-9}} \end{align*} {/eq}

The coefficient of {eq}x^0 {/eq} in the sum in orange will be the constant term in the given expansion. For that the power {eq}3k-9 {/eq} of {eq}x {/eq} should equal {eq}0 {/eq}. That is, {eq}3k-9=0 \Rightarrow k=3 {/eq}. So the constant term, that is, the coefficient of {eq}x^0 {/eq} is

{eq}\begin{align*} &\, \frac {9!}{k!(9-k)!}(3^k)(-1)^{9-k}\bigg|_{k=3}\\[2ex] =&\, \frac {9!}{3!(9-3)!}(3^3)(-1)^{9-3}\\[2ex] =&\, \frac {9!}{6*6!}(3^3)(-1)^{6}\\[2ex] =&\,\frac {9*8*7}{6}*27\\[2ex] =&\,\color{Blue}{2268\hspace{0.8cm}\mathrm{(Answer)}} \end{align*} {/eq}