1. **State the problem:** We need to find the possible values of $a$ such that the constant term in the expansion of $\left(x^{22} + a x\right)^6$ is 960. 2. **Recall the binomial expansion formula:** $$\left(u + v\right)^n = \sum_{k=0}^n \binom{n}{k} u^{n-k} v^k$$ 3. **Apply the formula to our expression:** Here, $u = x^{22}$, $v = a x$, and $n=6$. The general term is: $$T_{k+1} = \binom{6}{k} (x^{22})^{6-k} (a x)^k = \binom{6}{k} a^k x^{22(6-k)} x^k = \binom{6}{k} a^k x^{132 - 22k + k} = \binom{6}{k} a^k x^{132 - 21k}$$ 4. **Find the constant term:** The constant term corresponds to the power of $x$ being zero: $$132 - 21k = 0 \implies 21k = 132 \implies k = \frac{132}{21} = \frac{44}{7}$$ Since $k$ must be an integer between 0 and 6, there is no integer $k$ satisfying this. 5. **Re-examine the problem:** The problem states the constant term is 960, but from the powers of $x$, the exponent is $132 - 21k$. For the term to be constant, exponent must be zero, which is impossible. 6. **Check if the problem means the term independent of $x$ after simplification:** Since $x^{22}$ and $x$ are involved, the powers of $x$ in each term are $132 - 21k$. For $k=6$, exponent is $132 - 21*6 = 132 - 126 = 6$, not zero. 7. **Conclusion:** No term in the expansion is independent of $x$, so the constant term is zero unless $a=0$. **Therefore, the problem as stated has no solution for $a$ that yields a constant term of 960.**