Find the first three non-zero terms in the Maclaurin series for $e^{x}\sin(x+\pi)$. It is given that the three terms found in part (a) are equal to the first three terms in the series expansion of $a x (1 + b x)^{c}$ for small $x$, where $a$, $b$ and $c$ are constants. Find the exact values of $a$, $b$ and $c$. Use these values to find the coefficient of $x^{4}$ in the expansion of $a x (1 + b x)^{c}$, giving your answer as a simplified rational number.

1. Simplify the trigonometric part of the expression.

$\sin(x+\pi) = \sin x \cos \pi + \cos x \sin \pi = -\sin x$

Therefore, the expression becomes $-e^x \sin x$.

2. Write down the standard Maclaurin series expansions for $e^x$ and $\sin x$ up to the term in $x^3$.

$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots$
$\sin x = x - \frac{x^3}{6} + \dots$

Substituting these into our expression:
$-e^x \sin x = -\left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots\right)\left(x - \frac{x^3}{6} + \dots\right)$

3. Multiply the two series and collect the lowest three non-zero terms.

$-e^x \sin x = -\left( x - \frac{x^3}{6} + x^2 + \frac{x^3}{2} + \dots \right)$
$= -x - x^2 - \left(\frac{1}{2} - \frac{1}{6}\right)x^3 - \dots$
$= -x - x^2 - \frac{1}{3}x^3$

1. Determine the first three non-zero terms of $e^x\sin(x+\pi)$ (result from part a).

$\sin(x+\pi) = -\sin x$
Using standard expansions:
$e^x \sin(x+\pi) = -\left(1 + x + \frac{x^2}{2} + \dots\right)\left(x - \frac{x^3}{6} + \dots\right)$
$= -\left(x + x^2 + \left(\frac{1}{2} - \frac{1}{6}\right)x^3 + \dots\right)$
$= -x - x^2 - \frac{1}{3}x^3$

2. Expand $ax(1+bx)^c$ up to the term in $x^3$.

$ax(1+bx)^c = ax \left( 1 + cbx + \frac{c(c-1)}{2}(bx)^2 + \dots \right)$
$= ax + abcx^2 + \frac{ab^2c(c-1)}{2}x^3 + \dots$

3. Equate coefficients of corresponding powers of $x$ to form a system of equations.

Comparing $-x - x^2 - \frac{1}{3}x^3$ with $ax + abcx^2 + \frac{ab^2c(c-1)}{2}x^3$:
Coefficient of $x$: $a = -1$
Coefficient of $x^2$: $abc = -1$
Coefficient of $x^3$: $\frac{ab^2c(c-1)}{2} = -\frac{1}{3}$

4. Solve the system of equations for $b$ and $c$.

Substitute $a = -1$ into $abc = -1$:
$(-1)bc = -1 \implies bc = 1 \implies b = \frac{1}{c}$
Substitute $a = -1$ and $b = \frac{1}{c}$ into the third equation:
$\frac{-1}{2}\left(\frac{1}{c}\right)^2c(c-1) = -\frac{1}{3}$
$\frac{c-1}{2c} = \frac{1}{3}$
$3(c - 1) = 2c \implies 3c - 3 = 2c \implies c = 3$
Since $b = \frac{1}{c}$, we get $b = \frac{1}{3}$.

5. Find the coefficient of $x^4$ in the expansion using the exact values.

With $a = -1$, $b = \frac{1}{3}$, and $c = 3$, the expression is $-x\left(1+\frac{1}{3}x\right)^3$.
The $x^4$ term comes from multiplying $-x$ by the term in $x^3$ from the expansion of $\left(1+\frac{1}{3}x\right)^3$.
Term in $x^3$ is $\binom{3}{3}\left(\frac{1}{3}x\right)^3 = \frac{1}{27}x^3$.
$-x \left( \frac{1}{27}x^3 \right) = -\frac{1}{27}x^4$.
The coefficient of $x^4$ is $-\frac{1}{27}$.