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Step-by-Step Solution
Step 1: Identify the Binomial Expression
We are given the expression:
$ \left(a\,x^{\frac{1}{8}} + b\,x^{-\frac{1}{12}}\right)^{10} \,.$
Our goal is to find the minimum value of its constant term (the term independent of $x$) under the condition:
$ \frac{1}{a^2} + \frac{1}{b^3} = 4 \,.$
Step 2: Write the General Term of the Binomial Expansion
The general term in the binomial expansion of $ (P + Q)^{10} $ is given by:
$ T_{r+1} = {}^{10}C_{r}\,(P)^{10-r}\,(Q)^{r} \,.$
Here,
$P = a\,x^{\frac{1}{8}}$ and
$Q = b\,x^{-\frac{1}{12}}\,.$
Therefore, the general term is:
$$
T_{r+1}
= {}^{10}C_r \, \left(a\,x^{\frac{1}{8}}\right)^{10-r} \,\left(b\,x^{-\frac{1}{12}}\right)^{r}.
$$
Simplifying further:
$$
T_{r+1}
= {}^{10}C_r \, a^{\,10-r}\, b^{r} \, x^{\frac{10-r}{8}} \, x^{-\frac{r}{12}}
= {}^{10}C_r \, a^{\,10-r}\, b^{r} \, x^{\left(\frac{10-r}{8} - \frac{r}{12}\right)}.
$$
Combine the exponents of $x$:
$$
\frac{10 - r}{8} - \frac{r}{12}
= \frac{30 - 3r}{24} - \frac{2r}{24}
= \frac{30 - 5r}{24}.
$$
Thus:
$$
T_{r+1}
= {}^{10}C_r \, a^{\,10-r} \, b^{r} \, x^{\frac{30 - 5r}{24}}.
$$
Step 3: Find the Condition for the Constant Term
The constant term arises when the power of $x$ is zero, that is:
$$
\frac{30 - 5r}{24} = 0
\quad \Longrightarrow \quad
30 - 5r = 0
\quad \Longrightarrow \quad
r = 6.
$$
Hence, the constant term is $T_{6+1} = T_{7}.$
Step 4: Evaluate the Constant Term
Substitute $r=6$ into the general term:
$$
T_{7}
= {}^{10}C_{6} \, a^{\,10-6} \, b^{6}
= {}^{10}C_{6} \, a^{4} \, b^{6}.
$$
Compute the binomial coefficient:
$$
{}^{10}C_{6}
= \frac{10!}{6! \, 4!}
= \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}
= 210.
$$
Therefore,
$$
T_{7} = 210 \, a^{4} \, b^{6}.
$$
Step 5: Use the Given Condition to Find the Minimum of $a^4 b^6$
We know:
$$
\frac{1}{a^2} + \frac{1}{b^3} = 4,
$$
and we want to minimize $a^{4} b^{6}.$ We use the inequality involving harmonic mean (HM) and geometric mean (GM) for the two positive numbers $a^2$ and $b^3.$
The GM of $a^2$ and $b^3$ is:
$$
\sqrt{a^2 \, b^3} = (a^2 b^3)^{\tfrac{1}{2}},
$$
while their harmonic mean is:
$$
\mathrm{HM} = \frac{2}{\frac{1}{a^2} + \frac{1}{b^3}}.
$$
Since $\text{GM} \ge \text{HM},$ we get:
$$
\sqrt{a^2 b^3}
\;\ge\; \frac{2}{\frac{1}{a^2} + \frac{1}{b^3}}
= \frac{2}{4}
= \frac{1}{2}.
$$
Squaring both sides:
$$
a^2 b^3 \;\ge\; \frac{1}{4}
\quad \Longrightarrow \quad
(a^2 b^3)^2 = a^4 b^6 \;\ge\; \frac{1}{16}.
$$
This implies the minimum possible value of $a^4 b^6$ is $\frac{1}{16}.$
Step 6: Determine the Minimum Constant Term
Since $ T_{7} = 210 \, a^{4} \, b^{6}, $ the minimum value occurs when $a^4 b^6 = \frac{1}{16}.$ Therefore,
$$
\min(T_{7}) = 210 \times \frac{1}{16}
= \frac{210}{16}
= \frac{105}{8}.
$$
Final Answer
The minimum value of the constant term is
$ \boxed{\frac{105}{8}}.
$
