Question:

Step 1: Determine the general term for the first expansion. The general term in the binomial expansion of $(A+B)^n$ is given by $T_{r+1} = \binom{n}{r} A^{n-r} B^r$. For the first expression, $\left(\alpha x^3+\dfrac{1}{\beta x}\right)^{11}$, we have $A = \alpha x^3$, $B = \dfrac{1}{\beta x}$, and $n=11$. Thus, the general term is: $$T_{r+1} = \binom{11}{r} (\alpha x^3)^{11-r} \left(\frac{1}{\beta x}\right)^r$$ $$T_{r+1} = \binom{11}{r} \alpha^{11-r} (x^3)^{11-r} \beta^{-r} x^{-r}$$ $$T_{r+1} = \binom{11}{r} \alpha^{11-r} \beta^{-r} x^{3(11-r)-r}$$ $$T_{r+1} = \binom{11}{r} \alpha^{11-r} \beta^{-r} x^{33-3r-r}$$ $$T_{r+1} = \binom{11}{r} \alpha^{11-r} \beta^{-r} x^{33-4r}$$ Step 2: Find the value of $r$ for the coefficient of $x^9$ in the first expansion. To find the coefficient of $x^9$, we equate the power of $x$ in the general term to 9: $$33-4r = 9$$ $$4r = 33-9$$ $$4r = 24$$ $$r = 6$$ Step 3: Write down the coefficient of $x^9$ from the first expansion. Substitute $r=6$ into the coefficient part of the general term: $$\text{Coefficient of } x^9 = \binom{11}{6} \alpha^{11-6} \beta^{-6} = \binom{11}{6} \alpha^5 \beta^{-6}$$ Step 4: Determine the general term for the second expansion. For the second expression, $\left(\alpha x-\dfrac{1}{\beta x^3}\right)^{11}$, we have $A = \alpha x$, $B = -\dfrac{1}{\beta x^3}$, and $n=11$. Thus, the general term is: $$T'_{r+1} = \binom{11}{r} (\alpha x)^{11-r} \left(-\frac{1}{\beta x^3}\right)^r$$ $$T'_{r+1} = \binom{11}{r} \alpha^{11-r} x^{11-r} (-1)^r \beta^{-r} (x^{-3})^r$$ $$T'_{r+1} = \binom{11}{r} \alpha^{11-r} (-1)^r \beta^{-r} x^{11-r-3r}$$ $$T'_{r+1} = \binom{11}{r} \alpha^{11-r} (-1)^r \beta^{-r} x^{11-4r}$$ Step 5: Find the value of $r$ for the coefficient of $x^{-9}$ in the second expansion. To find the coefficient of $x^{-9}$, we equate the power of $x$ in the general term to -9: $$11-4r = -9$$ $$4r = 11+9$$ $$4r = 20$$ $$r = 5$$ Step 6: Write down the coefficient of $x^{-9}$ from the second expansion. Substitute $r=5$ into the coefficient part of the general term: $$\text{Coefficient of } x^{-9} = \binom{11}{5} \alpha^{11-5} (-1)^5 \beta^{-5}$$ $$\text{Coefficient of } x^{-9} = \binom{11}{5} \alpha^6 (-1) \beta^{-5} = -\binom{11}{5} \alpha^6 \beta^{-5}$$ Step 7: Equate the two coefficients. According to the problem statement, the coefficient of $x^9$ in the first expansion and the coefficient of $x^{-9}$ in the second expansion are equal: $$\binom{11}{6} \alpha^5 \beta^{-6} = -\binom{11}{5} \alpha^6 \beta^{-5}$$ We know that $\binom{n}{k} = \binom{n}{n-k}$, so $\binom{11}{6} = \binom{11}{11-6} = \binom{11}{5}$. Let $C = \binom{11}{5}$. The equation becomes: $$C \alpha^5 \beta^{-6} = -C \alpha^6 \beta^{-5}$$ Since $C \neq 0$, we can divide both sides by $C$: $$\alpha^5 \beta^{-6} = -\alpha^6 \beta^{-5}$$ Step 8: Solve for $(\alpha\beta)^2$. Divide both sides by $\alpha^5 \beta^{-5}$ (assuming $\alpha, \beta \neq 0$): $$\frac{\alpha^5 \beta^{-6}}{\alpha^5 \beta^{-5}} = -\frac{\alpha^6 \beta^{-5}}{\alpha^5 \beta^{-5}}$$ $$\beta^{-6 - (-5)} = -\alpha^{6-5}$$ $$\beta^{-1} = -\alpha^1$$ $$\frac{1}{\beta} = -\alpha$$ $$1 = -\alpha\beta$$ Square both sides of the equation $1 = -\alpha\beta$: $$(1)^2 = (-\alpha\beta)^2$$ $$1 = (\alpha\beta)^2$$ Step 9: State the final answer. The value of $(\alpha\beta)^2$ is 1. The final answer is $\boxed{1}$.