Question

find the term independent of x in the expansion of 3 by 2 x square minus 1 by 3 X whole to the power 6 ​

Answer 1

Step-by-step explanation:

$(\frac{3}{2} }x^{2} -\frac{1}{3}x )^{6}$

now we use general term of binomial theroem

$T_{r+1} =$ $^{6} C_{r}$ $( \frac{3}{2} x^{2})^{6-r}$$(\frac{-1}{3} x)^{r}$

    = $6c_{r} (\frac{3}{2})^{6-r} x^{12-3r}(\frac{-1}{3})^{r}$  ---------(1)

the term will be independent the whole power of x equal to zero

   ⇒12-3r=0

  ⇒r = 4    using r in equation (1)

  ⇒  $6c_{4 (\frac{3}{2}) ^{2}\frac{1}{(3)^{4} }} }$ now solving we get

  ⇒    $\frac{5}{12}$