Binomial Expansion rth term 
Given that there's no term on $x^3$ in $$(k + 2x) (1 - 3/2 x) ^6.$$ Evaluate the value of $k$ 

Difficult 
After expanding $(1 - 3/2 x)^6$ up to the 3rd term I get $1 - 9x + \frac{135}{2} x^2 - \frac{135}{4} x^3 + \ldots$ 
What do I do next to solve this question?? 

Hint
The terms with $x^3$ are
$$k\cdot \binom{6}{3}\cdot 1^3\cdot \left(-\frac 32\right)^3\cdot x^3$$ and
$$2x\cdot \binom{6}{4}\cdot 1^2\cdot \left(-\frac 32\right)^2\cdot x^2.$$
