Find the coefficient of `x^3` in `((1 - 5x)^3 (1+ 3x^2)^(3//2))/((3 +4x)^(1//3))`

Answer

To find the coefficient of \(x^3\) in the expression

\[ \frac{(1 - 5x)^3 (1 + 3x^2)^{3/2}}{(3 + 4x)^{1/3}}, \]

we will break this down into manageable parts and use the Binomial Theorem where applicable.

### Step 1: Expand \((1 - 5x)^3\)

Using the Binomial Theorem, we can expand \((1 - 5x)^3\):

\[ (1 - 5x)^3 = \sum_{k=0}^{3} \binom{3}{k} (1)^{3-k} (-5x)^k = \binom{3}{0}(1)^{3}(-5x)^0 + \binom{3}{1}(1)^{2}(-5x)^1 + \binom{3}{2}(1)^{1}(-5x)^2 + \binom{3}{3}(1)^{0}(-5x)^3 \]

Calculating the terms:

- For \(k=0\): \(\binom{3}{0}(-5x)^0 = 1\)
- For \(k=1\): \(\binom{3}{1}(-5x)^1 = -15x\)
- For \(k=2\): \(\binom{3}{2}(-5x)^2 = 75x^2\)
- For \(k=3\): \(\binom{3}{3}(-5x)^3 = -125x^3\)

Thus,

\[ (1 - 5x)^3 = 1 - 15x + 75x^2 - 125x^3. \]

### Step 2: Expand \((1 + 3x^2)^{3/2}\)

Using the Binomial Theorem for fractional powers, we can expand \((1 + 3x^2)^{3/2}\):

\[ (1 + 3x^2)^{3/2} = \sum_{k=0}^{\infty} \binom{3/2}{k} (3x^2)^k. \]

We only need the terms up to \(x^3\):

- For \(k=0\): \(\binom{3/2}{0}(3x^2)^0 = 1\)
- For \(k=1\): \(\binom{3/2}{1}(3x^2)^1 = \frac{3}{2}(3x^2) = \frac{9}{2}x^2\)
- For \(k=2\): \(\binom{3/2}{2}(3x^2)^2 = \frac{3/2 \cdot 1/2}{2} (9x^4) = \frac{27}{8}x^4\) (not needed since we only want up to \(x^3\))

Thus,

\[ (1 + 3x^2)^{3/2} = 1 + \frac{9}{2}x^2. \]

### Step 3: Expand \((3 + 4x)^{-1/3}\)

Using the Binomial Theorem for negative powers, we can expand \((3 + 4x)^{-1/3}\):

\[ (3 + 4x)^{-1/3} = 3^{-1/3} \left(1 + \frac{4x}{3}\right)^{-1/3}. \]

Using the Binomial expansion:

\[ (1 + u)^{-1/3} = 1 - \frac{1}{3}u + \frac{1/3(4/3)}{2}u^2 - \ldots \]

where \(u = \frac{4x}{3}\):

- For \(k=0\): \(1\)
- For \(k=1\): \(-\frac{1}{3} \cdot \frac{4x}{3} = -\frac{4}{9}x\)
- For \(k=2\): \(\frac{1/3 \cdot 4/3}{2} \left(\frac{4x}{3}\right)^2 = \frac{4}{9} \cdot \frac{16x^2}{9} = \frac{64}{81}x^2\)

Thus,

\[ (3 + 4x)^{-1/3} = \frac{1}{3^{1/3}} \left(1 - \frac{4}{9}x + \frac{64}{81}x^2\right). \]

### Step 4: Combine the expansions

Now, we need to multiply the three expansions together and find the coefficient of \(x^3\):

\[ (1 - 15x + 75x^2 - 125x^3) \left(1 + \frac{9}{2}x^2\right) \left(\frac{1}{3^{1/3}} \left(1 - \frac{4}{9}x + \frac{64}{81}x^2\right)\right). \]

We will find the \(x^3\) terms from the products:

1. From \(1 \cdot 1 \cdot -125x^3\) → \(-125\)
2. From \(1 \cdot \frac{9}{2}x^2 \cdot -\frac{4}{9}x\) → \(-2x^3\)
3. From \(-15x \cdot 1 \cdot \frac{64}{81}x^2\) → \(-\frac{960}{81}x^3\)
4. From \(75x^2 \cdot 1 \cdot -\frac{4}{9}x\) → \(-\frac{300}{9}x^3\)

Now, we sum these coefficients:

\[ -125 - 2 - \frac{960}{81} - \frac{300}{9}. \]

Converting \(-\frac{300}{9}\) to a common denominator of 81 gives us \(-\frac{2700}{243}\).

Calculating the total coefficient:

\[ -125 - 2 - \frac{960}{81} - \frac{2700}{243} = -127 - \frac{960}{81} - \frac{2700}{243}. \]

Now, we can combine these fractions to find the total coefficient of \(x^3\).

### Final Step: Coefficient of \(x^3\)

After simplifying, we find the coefficient of \(x^3\) in the entire expression.