1. Key Concept: The General Term of a Binomial Expansion

For a binomial expression of the form $(a+b)^n$, the general term (or the $(r+1)^{th}$ term) in its expansion is given by the formula: $T_{r+1} = {^nC_r} a^{n-r} b^r$ where:

• $n$ is the power to which the binomial is raised.
• $r$ is the index of the second term (starting from $r=0$ for the first term).
• ${^nC_r} = \frac{n!}{r!(n-r)!}$ is the binomial coefficient, representing the number of ways to choose $r$ items from a set of $n$ items.
• $a$ is the first term of the binomial.
• $b$ is the second term of the binomial.

This formula allows us to find any specific term in the expansion without writing out the entire expansion, which is particularly useful for finding terms with specific characteristics, like the constant term.

2. Applying the Binomial Theorem to the Given Expression

The given expression is $\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}$. Here, we have:

• $n = 12$
• $a = \frac{\sqrt[5]{3}}{x} = 3^{1/5} x^{-1}$
• $b = \frac{2 x}{\sqrt[3]{5}} = 2 \cdot x \cdot 5^{-1/3}$

Now, substitute these into the general term formula: $T_{r+1} = {^{12}C_r} \left(3^{1/5} x^{-1}\right)^{12-r} \left(2 \cdot x \cdot 5^{-1/3}\right)^r$

To simplify, we distribute the exponents and group terms with $x$: $T_{r+1} = {^{12}C_r} (3^{1/5})^{12-r} (x^{-1})^{12-r} (2)^r (x)^r (5^{-1/3})^r$ $T_{r+1} = {^{12}C_r} \cdot 3^{\frac{12-r}{5}} \cdot x^{-(12-r)} \cdot 2^r \cdot x^r \cdot 5^{-\frac{r}{3}}$

Now, combine the powers of $x$: $T_{r+1} = {^{12}C_r} \cdot 3^{\frac{12-r}{5}} \cdot 2^r \cdot 5^{-\frac{r}{3}} \cdot x^{(-12+r+r)}$ $T_{r+1} = {^{12}C_r} \cdot 3^{\frac{12-r}{5}} \cdot 2^r \cdot 5^{-\frac{r}{3}} \cdot x^{2r-12}$

3. Finding the Term Independent of $x$ (Constant Term)

A constant term in a binomial expansion is a term that does not contain the variable $x$. For this to happen, the exponent of $x$ in the general term must be equal to zero.

Set the exponent of $x$ to zero: $2r - 12 = 0$ Solve for $r$: $2r = 12$ $r = 6$

This means the constant term is the $(6+1)^{th}$, or the $7^{th}$ term, in the expansion.

4. Calculating the Constant Term

Now, substitute $r=6$ back into the general term expression for the coefficients (excluding the $x$ term, as its power is now 0): Constant term $T_{6+1} = {^{12}C_6} \cdot 3^{\frac{12-6}{5}} \cdot 2^6 \cdot 5^{-\frac{6}{3}}$

Let's calculate each part:

• Binomial Coefficient: ${^{12}C_6} = \frac{12!}{6!6!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 11 \times 2 \times 3 \times 2 \times 7 = 924$
• Power of 3: $3^{\frac{12-6}{5}} = 3^{\frac{6}{5}} = 3^{1 + 1/5} = 3 \cdot 3^{1/5} = 3\sqrt[5]{3}$
• Power of 2: $2^6 = 64$
• Power of 5: $5^{-\frac{6}{3}} = 5^{-2} = \frac{1}{5^2} = \frac{1}{25}$

Now, multiply these values together: Constant term $= 924 \times (3\sqrt[5]{3}) \times 64 \times \frac{1}{25}$ Constant term $= \frac{924 \times 3 \times 64}{25} \sqrt[5]{3}$ Constant term $= \frac{177408}{25} \sqrt[5]{3}$

5. Comparing with the Given Form and Finding $\alpha$

The problem states that the constant term is $\alpha \times 2^8 \times \sqrt[5]{3}$. We need to express our calculated constant term in this format. Notice that $2^8 = 2^6 \times 2^2 = 64 \times 4 = 256$.

Our calculated constant term is $\frac{924 \times 3 \times 64}{25} \sqrt[5]{3}$. We want a factor of $2^8$ (which is $256$). We can factor $64$ out directly, but we need an additional factor of $4$ from $924$. $924 = 4 \times 231$

So, the constant term can be written as: Constant term $= \frac{(4 \times 231) \times 3 \times 64}{25} \sqrt[5]{3}$ Constant term $= \frac{231 \times 3 \times (4 \times 64)}{25} \sqrt[5]{3}$ Constant term $= \frac{231 \times 3 \times 256}{25} \sqrt[5]{3}$ Constant term $= \frac{693 \times 2^8}{25} \sqrt[5]{3}$

Now, compare this with the given form $\alpha \times 2^8 \times \sqrt[5]{3}$: $\frac{693}{25} \times 2^8 \times \sqrt[5]{3} = \alpha \times 2^8 \times \sqrt[5]{3}$

By comparing the coefficients, we find: $\alpha = \frac{693}{25}$

6. Final Calculation: $25\alpha$

The question asks for the value of $25\alpha$. $25\alpha = 25 \times \frac{693}{25}$ $25\alpha = 693$

7. Tips for Success and Common Mistakes

• Careful with Exponents: Pay close attention to negative exponents and fractional exponents, especially when dealing with variables in the denominator or roots.
• Radical to Power Form: Convert roots (like $\sqrt[5]{3}$) to fractional exponents ($3^{1/5}$) for easier manipulation during calculations.
• Variable Grouping: When finding the term independent of $x$, ensure all $x$ terms are grouped correctly and their powers are summed algebraically. A common mistake is to forget a negative sign or incorrectly combine powers.
• Arithmetic Precision: Binomial coefficient calculations and subsequent multiplications can be large. Use a calculator for large numbers if allowed, or simplify carefully step-by-step to avoid errors.
• Matching Form: When comparing your result to a given form (like $\alpha \times 2^8 \times \sqrt[5]{3}$), be prepared to factor or rearrange your terms to match the required structure. This might involve splitting numbers into their prime factors or powers of a specific base.

8. Summary

This problem effectively tests the application of the binomial theorem's general term formula. The key steps involved setting up the general term, isolating the variable component to find the condition for a constant term, calculating the value of $r$, and then substituting $r$ back into the formula to find the numerical constant. Finally, careful algebraic manipulation was required to match the given form of the constant term and solve for $\alpha$. The value of $25\alpha$ was found to be $693$.

The final answer is $\boxed{\text{693}}$.