"no terms independent of x" basically means there is no constant term, or the coefficient of x ⁰ is zero.
Recall the binomial theorem:
[tex]\displaystyle (a+b)^n=\sum_{k=0}^n\binom nk a^{n-k}b^k[/tex]
So we have
[tex]\displaystyle \left(1+ax^2\right)\left(\frac2x-3x\right)^6 = \left(1+ax^2\right) \sum_{k=0}^6 \binom 6k \left(\frac2x\right)^{6-k}(-3x)^k[/tex]
[tex]\displaystyle \left(1+ax^2\right)\left(\frac2x-3x\right)^6 = \left(1+ax^2\right) \sum_{k=0}^6 2^{6-k}(-3)^k\binom 6k x^{2k-6}[/tex]
[tex]\displaystyle \left(1+ax^2\right)\left(\frac2x-3x\right)^6 = \sum_{k=0}^6 2^{6-k}(-3)^k\binom 6k x^{2k-6}+a\sum_{k=0}^6 2^{6-k}(-3)^k\binom 6k x^{2k-4}[/tex]
The first sum contributes a x ⁰ term for 2k - 6 = 0, or k = 3, while the second sum contributes a x ⁰ term for 2k - 4 = 0, or k = 2. The coefficient of the sum of these terms must be zero:
[tex]2^{6-3}(-3)^3\dbinom 63 + a\times2^{6-2}(-3)^2\dbinom 62 = 0[/tex]
which reduces to
2160a - 4320 = 0
2160a = 4320
a = 2
