Key Concept: The Binomial Theorem

The general term, $T_{r+1}$, in the binomial expansion of $(a+b)^n$ is given by: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ where $n$ is the power of the binomial, $r$ is the index of the term (starting from 0), $a$ is the first term, and $b$ is the second term. This formula is fundamental for extracting specific terms or their coefficients from a binomial expansion without having to expand the entire expression.

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Step-by-Step Solution

1. Finding the Coefficient of $x^9$ in ${\left( {\alpha {x^3} + {1 \over {\beta x}}} \right)^{11}}$

We begin by analyzing the first expression. Here, we have $n=11$, the first term $a = \alpha x^3$, and the second term $b = \frac{1}{\beta x} = \frac{1}{\beta} x^{-1}$.

Using the general term formula, $T_{r+1} = \binom{n}{r} a^{n-r} b^r$: $T_{r+1} = \binom{11}{r} (\alpha x^3)^{11-r} \left(\frac{1}{\beta x}\right)^r$

To find the coefficient of $x^9$, we need to isolate the powers of $x$. We do this by distributing the exponents and combining the $x$ terms: $T_{r+1} = \binom{11}{r} \alpha^{11-r} (x^3)^{11-r} \left(\frac{1}{\beta}\right)^r (x^{-1})^r$ $T_{r+1} = \binom{11}{r} \frac{\alpha^{11-r}}{\beta^r} x^{3(11-r)} x^{-r}$ $T_{r+1} = \binom{11}{r} \frac{\alpha^{11-r}}{\beta^r} x^{33-3r-r}$ $T_{r+1} = \binom{11}{r} \frac{\alpha^{11-r}}{\beta^r} x^{33-4r}$

Now, we set the exponent of $x$ to 9 to find the value of $r$ that corresponds to the $x^9$ term: $33 - 4r = 9$ $4r = 33 - 9$ $4r = 24$ $r = 6$

Substituting $r=6$ back into the coefficient part of $T_{r+1}$ gives us the desired coefficient: $\text{Coefficient of } x^9 = \binom{11}{6} \frac{\alpha^{11-6}}{\beta^6} = \binom{11}{6} \frac{\alpha^5}{\beta^6}$

2. Finding the Coefficient of $x^{-9}$ in ${\left( {\alpha x - {1 \over {\beta {x^3}}}} \right)^{11}}$

Next, we analyze the second expression. Here, $n=11$, the first term $a = \alpha x$, and the second term $b = - \frac{1}{\beta x^3} = - \frac{1}{\beta} x^{-3}$. It is crucial to include the negative sign with the second term, as forgetting it is a common error.

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

Again, we separate coefficients and powers of $x$: $T_{r+1} = \binom{11}{r} \alpha^{11-r} x^{11-r} \left(- \frac{1}{\beta}\right)^r (x^{-3})^r$ $T_{r+1} = \binom{11}{r} (-1)^r \frac{\alpha^{11-r}}{\beta^r} x^{11-r} x^{-3r}$ $T_{r+1} = \binom{11}{r} (-1)^r \frac{\alpha^{11-r}}{\beta^r} x^{11-r-3r}$ $T_{r+1} = \binom{11}{r} (-1)^r \frac{\alpha^{11-r}}{\beta^r} x^{11-4r}$

We are looking for the term with $x^{-9}$, so we set the exponent of $x$ to -9: $11 - 4r = -9$ $4r = 11 + 9$ $4r = 20$ $r = 5$

Substituting $r=5$ back into the coefficient part of $T_{r+1}$: $\text{Coefficient of } x^{-9} = \binom{11}{5} (-1)^5 \frac{\alpha^{11-5}}{\beta^5}$ Since $(-1)^5 = -1$: $\text{Coefficient of } x^{-9} = - \binom{11}{5} \frac{\alpha^6}{\beta^5}$

3. Equating the Coefficients and Solving for $(\alpha\beta)^2$

The problem states that these two coefficients are equal: $\binom{11}{6} \frac{\alpha^5}{\beta^6} = - \binom{11}{5} \frac{\alpha^6}{\beta^5}$

We use the property of binomial coefficients: $\binom{n}{r} = \binom{n}{n-r}$. Thus, $\binom{11}{6} = \binom{11}{11-6} = \binom{11}{5}$. Since both $\binom{11}{6}$ and $\binom{11}{5}$ are equal and non-zero, we can cancel them from both sides of the equation: $\frac{\alpha^5}{\beta^6} = - \frac{\alpha^6}{\beta^5}$

Assuming $\alpha \ne 0$ and $\beta \ne 0$ (otherwise, the original terms would be undefined or trivial), we can simplify by dividing both sides by $\alpha^5$ and multiplying both sides by $\beta^6$: $\frac{1}{\beta^6} \cdot \beta^6 = - \frac{\alpha^6}{\beta^5} \cdot \frac{\beta^6}{\alpha^5}$ $1 = - \alpha^{6-5} \beta^{6-5}$ $1 = - \alpha \beta$ $\alpha \beta = -1$

Finally, we need to find the value of $(\alpha\beta)^2$: $(\alpha\beta)^2 = (-1)^2$ $(\alpha\beta)^2 = 1$

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Relevant Tips and Common Mistakes:

• Sign Errors: The most frequent mistake in such problems is mishandling the negative sign when the second term of the binomial is negative (e.g., $(a-b)^n$). Always treat $b$ as the complete second term, including its sign.
• Exponent Rules: Be meticulous when combining powers of $x$, especially with negative exponents. Recall that $x^m \cdot x^n = x^{m+n}$ and $(x^m)^n = x^{mn}$.
• Binomial Coefficient Symmetry: The property $\binom{n}{r} = \binom{n}{n-r}$ is very useful for simplifying equations and recognizing equalities between binomial coefficients.
• Assumptions: When cancelling terms involving variables (like $\alpha^5$ and $\beta^5$), it's implicitly assumed that these variables are non-zero. In competitive exams, this assumption is generally safe unless specified otherwise.

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Summary and Key Takeaway

This problem is a good application of the Binomial Theorem, specifically finding particular terms in an expansion. The core idea is to express the general term, extract the power of the variable, solve for the term index $r$, and then use this $r$ to find the complete coefficient. Careful handling of negative signs and exponent rules, along with knowing binomial coefficient properties, are vital for accuracy. The problem culminates in solving a simple algebraic equation to find the final value.