Answer:
To find the term independent of \( x \) in the expansion of \( (3/2x^2 - 1/3x)^6 \), we need to find the term where the power of \( x \) is zero.
Using the binomial theorem, the general term of the expansion is given by:
\[ T_r = \binom{n}{r} (3/2x^2)^{n-r} (-1/3x)^r \]
For the term to be independent of \( x \), the powers of \( x \) in both terms \( (3/2x^2)^{n-r} \) and \( (-1/3x)^r \) must cancel each other out. So, we need:
\[ 2(n-r) - 1r = 0 \]
\[ 2n - 2r - r = 0 \]
\[ 2n - 3r = 0 \]
\[ r = \frac{2n}{3} \]
Since \( r \) must be a whole number, the largest whole number less than or equal to \( \frac{2n}{3} \) is \( \frac{2n}{3} = 2 \).
This implies \( n = 3 \).
Therefore, the term independent of \( x \) occurs when \( r = 2 \).
Substitute \( n = 3 \) and \( r = 2 \) into the general term:
\[ T_2 = \binom{3}{2} (3/2x^2)^{3-2} (-1/3x)^2 \]
\[ T_2 = 3 \times \frac{1}{2} \times (3/2x^2) (-1/3x)^2 \]
\[ T_2 = \frac{3}{2} \times \frac{9}{4}x^4 \times \frac{1}{9}x^2 \]
\[ T_2 = \frac{27}{8}x^6 \]
Therefore, the term independent of \( x \) in the expansion of \( (3/2x^2 - 1/3x)^6 \) is \( \frac{27}{8} \).
