Find the term INDEPENDENT of $x$ in the expansion of $\left(x^2 + \dfrac{3}{x}\right)^6$.

1. General term: $T_{r+1} = \binom{6}{r}\,(x^2)^{6-r}\,\left(\dfrac{3}{x}\right)^r = \binom{6}{r}\,3^r\,x^{12-2r-r} = \binom{6}{r}\,3^r\,x^{12-3r}$. 2. For the term INDEPENDENT of $x$, the power of $x$ must equal zero: $12 - 3r = 0 \Rightarrow r = 4$. 3. Substitute $r = 4$: term = $\binom{6}{4}\,3^4 = 15 \cdot 81 = 1215$. 4. So the term independent of $x$ is $1215$. 5. Option A uses $r = 3$ ($\binom{6}{3} \cdot 27 = 540$, wrong $r$). Option C is $\binom{6}{2} \cdot 9 = 135$ (also wrong $r$). Option D is double the answer. _Source: syllabus Mathematics, Ch 7, Example 1 (Similar method on $(x^2 + 3/x)^4$), p. 6._