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Step-by-step Solution
Step 1: Write the general term in the binomial expansion
For the expansion of $(x + y)^{n}$, the $(r+1)$-th term (counting from 1) is given by
$$
T_{r+1} = {}^{n}C_{r}\, x^{\,n-r}\,y^{\,r}.
$$
Here, $n=10$ and we have:
‣ $x = 2^{\frac{1}{3}}, \quad y = \frac{1}{2\,\bigl(3^{\frac{1}{3}}\bigr)}.$
Step 2: Identify the 5th term from the beginning
The 5th term from the beginning corresponds to $r=4$ because $T_{r+1}$ is the $(r+1)$-th term. Thus:
$$
T_{5}
= {}^{10}C_{4}\,\Bigl(2^{\tfrac{1}{3}}\Bigr)^{\,10-4}\,\Bigl(\tfrac{1}{2\,\bigl(3^{\tfrac{1}{3}}\bigr)}}\Bigr)^{4}.
$$
Step 3: Identify the 5th term from the end
In an expansion of $(x + y)^{10}$, there are 11 terms total ($T_{1}$ to $T_{11}$). The 5th term from the end is the $(11 - 5 + 1) = 7$-th term from the beginning. So, for $T_{7}$ we have $r=6$. However, the solution provided uses an equivalent ratio approach by pairing terms in a way that the binomial coefficients match (both ${}^{10}C_{4}$) and effectively inverts $x$ and $y$ exponents. Following the given structure in the solution:
We treat "5th from the end" in a form that gives:
$$
T_{5}^{(end)}
= {}^{10}C_{4}\,\Bigl(\tfrac{1}{2\,\bigl(3^{\tfrac{1}{3}}\bigr)}}\Bigr)^{\,10-4}\,\Bigl(2^{\tfrac{1}{3}}\Bigr)^{4}.
$$
Step 4: Form the ratio of the two terms
According to the given solution structure,
$$
\frac{T_{5}\bigl(\text{from beginning}\bigr)}{T_{5}\bigl(\text{from end}\bigr)}
= \frac{{}^{10}C_{4}\,\Bigl(2^{\tfrac{1}{3}}\Bigr)^{6}\,\Bigl(\tfrac{1}{2\,3^{\tfrac{1}{3}}}\Bigr)^{4}}
{{}^{10}C_{4}\,\Bigl(\tfrac{1}{2\,3^{\tfrac{1}{3}}}\Bigr)^{6}\,\Bigl(2^{\tfrac{1}{3}}\Bigr)^{4}}
= \frac{\Bigl(2^{\tfrac{1}{3}}\Bigr)^{6}}{\Bigl(2^{\tfrac{1}{3}}\Bigr)^{4}} \,\times
\frac{\Bigl(\tfrac{1}{2\,3^{\tfrac{1}{3}}}\Bigr)^{4}}{\Bigl(\tfrac{1}{2\,3^{\tfrac{1}{3}}}\Bigr)^{6}}.
$$
Step 5: Simplify the ratio
1. Combine the powers of $2^{\tfrac{1}{3}}$:
$$
\frac{\Bigl(2^{\tfrac{1}{3}}\Bigr)^{6}}{\Bigl(2^{\tfrac{1}{3}}\Bigr)^{4}}
= \Bigl(2^{\tfrac{1}{3}}\Bigr)^{6-4}
= 2^{\tfrac{2}{3}}.
$$
2. Combine the powers of $\bigl(\tfrac{1}{2\,3^{\tfrac{1}{3}}}\bigr)$:
$$
\frac{\Bigl(\tfrac{1}{2\,3^{\tfrac{1}{3}}}\Bigr)^{4}}{\Bigl(\tfrac{1}{2\,3^{\tfrac{1}{3}}}\Bigr)^{6}}
= \Bigl(\tfrac{1}{2\,3^{\tfrac{1}{3}}}\Bigr)^{-2}
= \Bigl(2\,3^{\tfrac{1}{3}}\Bigr)^{2}
= 4\,(3^{\tfrac{1}{3}})^{2}
= 4\,(3^{\tfrac{2}{3}}).
$$
3. Multiply these two results:
$$
2^{\tfrac{2}{3}} \times 4\,3^{\tfrac{2}{3}}
= 4\,\bigl(2^{\tfrac{2}{3}}\,3^{\tfrac{2}{3}}\bigr)
= 4\,\bigl((2\times 3)^{\tfrac{2}{3}}\bigr)
= 4\,(6^{\tfrac{2}{3}}).
$$
Since $\,(6^{\tfrac{2}{3}}) = \bigl((6^{\tfrac{1}{3}})^2\bigr)\,$ can be written as $(36)^{\tfrac{1}{3}}$, we have:
$$
4\,(6^{\tfrac{2}{3}})
= 4\,(36^{\tfrac{1}{3}}).
$$
Thus, the ratio is
$$
4\,(36)^{\tfrac{1}{3}} : 1.
$$
Final Answer:
$\displaystyle 4\,(36)^{\tfrac{1}{3}} : 1$
