Find the term independent of x in the expansion of a given binomial

Formula to use
$r^{th} \text{ term} = ({^nC_m})(a^{n - m})(b^m)$
 

For (x - 2/x)⁶:

a = x
b = -2/x
n = 6

 

$r^{th} \text{ term} = ({^6C_m})(x^{6 - m})\left( -\dfrac{2}{x} \right)^m$

$r^{th} \text{ term} = ({^6C_m})(-2)^m (x^{6 - 2m})$
 

Let $({^6C_m})(-2)^m = K_j$
 

For the r^th term involving K₁x⁰:

$6 - 2m = 0$

$m = 3$
 

$K_1 = ({^6C_3})(-2)^3 = -160$

 

For the r^th term involving K₂x²:

$6 - 2m = 2$

$m = 2$
 

$K_2 = ({^6C_2})(-2)^2 = 60$

 

The constant term in the expansion of (2 + 3/x²)(x - 2/x)⁶ is:
$K = 2K_1 + 3 K_2 = 2(-160) + 3(60)$

$K = -140$   ←   answer