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Step-by-Step Solution
Step 1: Identify the middle term in the expansion of $(1 + \alpha x)^4$
For a binomial expansion of $(1 + \alpha x)^n$, the general term (the $(k+1)$-th term) is given by
$${}^nC_k (\alpha x)^k.$$
When $n=4$, the middle term corresponds to the term where $k = \frac{n}{2} = 2$. Thus, the middle term is:
$${}^4C_2 (\alpha x)^2.$$
Therefore, its coefficient (the factor of $x^2$) is:
$${}^4C_2 \alpha^2.$$
We know that
$${}^4C_2 = 6.$$
So the coefficient of the middle term in $(1 + \alpha x)^4$ is
$$6\,\alpha^2.$$
Step 2: Identify the middle term in the expansion of $(1 - \alpha x)^6$
Similarly, for $(1 - \alpha x)^n$, the general term is
$${}^nC_k (-\alpha x)^k.$$
When $n=6$, the middle term corresponds to $k = \frac{n}{2} = 3$. Thus, the middle term is:
$${}^6C_3 (-\alpha x)^3.$$
Therefore, its coefficient (the factor of $x^3$) is:
$${}^6C_3(-\alpha)^3.$$
We know that
$${}^6C_3 = 20,$$
and $(-\alpha)^3 = -\alpha^3$. Hence, the coefficient of the middle term in $(1 - \alpha x)^6$ is
$$20 \cdot (-\alpha^3) = -20\,\alpha^3.$$
Step 3: Equate the two coefficients
According to the problem statement, these two coefficients are equal:
$$6\,\alpha^2 = -20\,\alpha^3.$$
Step 4: Solve for $\alpha$
Dividing both sides by $\alpha^2$ (assuming $\alpha \neq 0$), we get:
$$6 = -20\,\alpha.$$
Hence,
$$\alpha = -\frac{6}{20} = -\frac{3}{10}.$$
Answer
The value of $\alpha$ that makes the coefficients of the middle terms equal is
$$-\frac{3}{10}.$$
