Find the term independent of x in `((3)/(root(3)(x)) + 5 sqrtx)^25`

Answer

To find the term independent of \( x \) in the expression

\[ \left( \frac{3}{\sqrt[3]{x}} + 5\sqrt{x} \right)^{25}, \]

we will follow these steps:

### Step 1: Write the General Term
The general term \( T_r \) in the binomial expansion of \( (a + b)^n \) is given by:

\[ T_r = \binom{n}{r} a^{n-r} b^r. \]

In our case, \( a = \frac{3}{\sqrt[3]{x}} \), \( b = 5\sqrt{x} \), and \( n = 25 \). Thus, the general term can be expressed as:

\[ T_r = \binom{25}{r} \left( \frac{3}{\sqrt[3]{x}} \right)^{25-r} \left( 5\sqrt{x} \right)^r. \]

### Step 2: Simplify the General Term
We can rewrite the general term as:

\[ T_r = \binom{25}{r} \cdot 3^{25-r} \cdot \left( \frac{1}{x^{1/3}} \right)^{25-r} \cdot 5^r \cdot (x^{1/2})^r. \]

This simplifies to:

\[ T_r = \binom{25}{r} \cdot 3^{25-r} \cdot 5^r \cdot x^{-\frac{25-r}{3} + \frac{r}{2}}. \]

### Step 3: Find the Power of \( x \)
To find the term independent of \( x \), we need to set the exponent of \( x \) to zero:

\[ -\frac{25 - r}{3} + \frac{r}{2} = 0. \]

### Step 4: Solve for \( r \)
Multiplying the entire equation by 6 (the least common multiple of 3 and 2) to eliminate the fractions:

\[ -2(25 - r) + 3r = 0. \]

Expanding this gives:

\[ -50 + 2r + 3r = 0 \implies 5r = 50 \implies r = 10. \]

### Step 5: Find the Independent Term
Now, we substitute \( r = 10 \) back into the general term to find the term independent of \( x \):

\[ T_{10} = \binom{25}{10} \cdot 3^{25-10} \cdot 5^{10}. \]

Calculating this gives:

\[ T_{10} = \binom{25}{10} \cdot 3^{15} \cdot 5^{10}. \]

### Final Answer
Thus, the term independent of \( x \) is:

\[ \binom{25}{10} \cdot 3^{15} \cdot 5^{10}. \]