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Step 1: Write down the binomial expression
We need to analyze the expansion of
\left(\frac{x}{4} \;-\; \frac{12}{x^2}\right)^{12} .
Our objective is to find the term in this expansion that does not contain any power of x (i.e., the term independent of x ).
Step 2: Formulate the general term in the expansion
The general term T_{r+1} of a binomial expression
(A + B)^{n}
is given by
T_{r+1} = \binom{n}{r} A^{n-r} B^{r} .
In this problem:
A = \frac{x}{4}
B = -\frac{12}{x^2}
n = 12
Hence,
\[
T_{r+1} \;=\; \binom{12}{r} \left(\frac{x}{4}\right)^{12-r} \left(-\frac{12}{x^2}\right)^{r}.
\]
Step 3: Simplify the general term
First, factor out the sign and coefficients:
\[
T_{r+1}
= \binom{12}{r} \left(\frac{1}{4}\right)^{\,12-r} (x)^{\,12-r} \left(-1\right)^{r} \left(12\right)^{r} \left(\frac{1}{x^2}\right)^{r}.
\]
Combine the powers of x :
\[
(x)^{\,12-r} \times \left(\frac{1}{x^2}\right)^{r} = x^{\,12-r - 2r} = x^{\,12 - 3r}.
\]
Therefore,
\[
T_{r+1}
= (-1)^{r} \binom{12}{r} \left(\frac{1}{4}\right)^{\,12-r} (12)^{r} \, x^{\,12 - 3r}.
\]
Step 4: Impose the condition for the term independent of x
For T_{r+1} to be independent of x , the power of x must be 0:
\[
12 - 3r \;=\; 0 \;\;\Longrightarrow\;\; r = 4.
\]
Step 5: Substitute r=4 back into the general term
When r = 4 ,
\[
T_{5}
= (-1)^{4} \binom{12}{4} \left(\frac{1}{4}\right)^{8} \, (12)^{4}.
\]
Since (-1)^{4} = 1 , we have
\[
T_{5}
= \binom{12}{4} \left(\frac{1}{4}\right)^{8} \, 12^{4}.
\]
Step 6: Compare with the given form
We are told that the constant term appears in the form
\left(\frac{3^6}{4^4}\right) k .
We need to match
\[
\binom{12}{4} \left(\frac{1}{4}\right)^{8} \, 12^{4}
\quad \text{with} \quad
\frac{3^6}{4^4} \, k.
\]
First, let us simplify:
1. Factor out key components from 12^{4} = (3 \times 4)^{4} = 3^{4} \times 4^{4} .
2. Note that \left(\frac{1}{4}\right)^{8} = \frac{1}{4^{8}} .
So,
\[
\binom{12}{4} \left(\frac{1}{4^{8}}\right) \, (3^4 \times 4^4)
= \binom{12}{4} \left(\frac{3^4 \times 4^4}{4^{8}}\right)
= \binom{12}{4} \left(\frac{3^4}{4^4}\right).
\]
Hence, the term becomes
\[
\binom{12}{4} \frac{3^4}{4^4}.
\]
We want it to match
\[
\frac{3^6}{4^4} \, k.
\]
Notice we can write 3^4 as 3^6 \div 3^2 , so
\[
\binom{12}{4} \frac{3^4}{4^4}
= \binom{12}{4} \frac{3^6}{4^4} \cdot \frac{1}{3^2}
= \binom{12}{4} \frac{3^6}{4^4} \cdot \frac{1}{9}.
\]
Therefore, to match
\frac{3^6}{4^4} \, k , we can see that
\[
k
= \binom{12}{4} \cdot \frac{1}{9}.
\]
We compute
\[
\binom{12}{4} = \frac{12!}{4!\,8!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495.
\]
Then
\[
k = 495 \times \frac{1}{9} = 55.
\]
Step 7: State the final answer
The value of k is
55.
