Find the coefficient of $x^9$ in $\left(\alpha x^3+\frac{1}{\beta x}\right)^{11}$

The general term is $T_{r+1}=\binom{11}{r}(\alpha x^3)^{11-r}\left(\frac{1}{\beta x}\right)^r$

Simplifying, $T_{r+1}=\binom{11}{r}\alpha^{11-r}\beta^{-r}x^{3(11-r)-r}$ $=\binom{11}{r}\alpha^{11-r}\beta^{-r}x^{33-4r}$

For the power of $x$ to be $9$, $33-4r=9$ $4r=24$ $r=6$

So the coefficient of $x^9$ is $\binom{11}{6}\alpha^{5}\beta^{-6}$

Find the coefficient of $x^{-9}$ in $\left(\alpha x-\frac{1}{\beta x^3}\right)^{11}$

The general term is $T_{r+1}=\binom{11}{r}(\alpha x)^{11-r}\left(-\frac{1}{\beta x^3}\right)^r$

Simplifying, $T_{r+1}=\binom{11}{r}\alpha^{11-r}(-1)^r\beta^{-r}x^{(11-r)-3r}$ $=\binom{11}{r}\alpha^{11-r}(-1)^r\beta^{-r}x^{11-4r}$

For the power of $x$ to be $-9$, $11-4r=-9$ $4r=20$ $r=5$

Hence the coefficient of $x^{-9}$ is $\binom{11}{5}\alpha^{6}(-1)^5\beta^{-5}=-\binom{11}{5}\alpha^6\beta^{-5}$

Use the given condition that these coefficients are equal

Since $\binom{11}{6}=\binom{11}{5}$, we get $\binom{11}{6}\alpha^5\beta^{-6}=-\binom{11}{5}\alpha^6\beta^{-5}$ $\alpha^5\beta^{-6}=-\alpha^6\beta^{-5}$

Multiply both sides by $\beta^6$: $\alpha^5=-\alpha^6\beta$

Assuming the coefficients are nonzero, divide by $\alpha^5$: $1=-\alpha\beta$ $\alpha\beta=-1$

Therefore, $(\alpha\beta)^2 = (-1)^2=1$

Comparison with stored answer

Derived answer is $1$, which matches the stored correct answer.