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 $ax(1+bx)^{c}$ 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 $ax(1+bx)^{c}$, giving your answer as a simplified rational number.

1. Simplify the trigonometric part of the expression using compound angle or supplementary angle properties.

$\sin(x+\pi) = -\sin x$
$\Rightarrow e^x\sin(x+\pi) = -e^x\sin x$

2. State or substitute the standard Maclaurin series for $e^x$ and $\sin x$.

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

3. Multiply the polynomial expansions up to the required degree to obtain the first three non-zero terms.

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

1. Determine the first three non-zero terms of the Maclaurin series for $e^x\sin(x+\pi)$ (required from part (a) to solve part (b)).

Using the trigonometric identity $\sin(x+\pi) = -\sin x$:
$e^x \sin(x+\pi) = -e^x \sin x$
Substituting the standard Maclaurin series for $e^x$ and $\sin x$:
$-e^x \sin x = -\left(1 + x + \frac{x^2}{2} + \dots\right)\left(x - \frac{x^3}{6} + \dots\right)$
$= -\left(x - \frac{x^3}{6} + x^2 + \frac{x^3}{2} + \dots\right)$
$= -x - x^2 - \frac{1}{3}x^3$

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

Using the binomial expansion for $(1+bx)^c$:
$ax(1+bx)^c = ax \left( 1 + cbx + \frac{c(c-1)}{2!}(bx)^2 + \dots \right)$
$= ax + abcx^2 + \frac{ac(c-1)b^2}{2}x^3 + \dots$

3. Equate the coefficients of $x$, $x^2$, and $x^3$ from both series to set up a system of equations.

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

4. Solve the system of equations for the exact values of $b$ 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 $x^3$ coefficient equation:
$\frac{(-1)c(c-1)}{2} \left(\frac{1}{c}\right)^2 = -\frac{1}{3}$
$\frac{c-1}{2c} = \frac{1}{3}$
$3(c - 1) = 2c$
$3c - 3 = 2c \implies c = 3$
Since $bc = 1$, we get:
$b(3) = 1 \implies b = \frac{1}{3}$

5. Use the exact values of $a$, $b$, and $c$ to find the coefficient of $x^4$ in the expansion of $ax(1+bx)^c$.

Substitute $a = -1$, $b = \frac{1}{3}$, and $c = 3$ into the original expression:
$ax(1+bx)^c = -x\left(1 + \frac{1}{3}x\right)^3$
The term in $x^4$ comes from the $x^3$ term in the expansion of $\left(1 + \frac{1}{3}x\right)^3$ multiplied by $-x$:
$-x \times \binom{3}{3}\left(\frac{1}{3}x\right)^3 = -x \left(\frac{1}{27}x^3\right) = -\frac{1}{27}x^4$
Therefore, the coefficient of $x^4$ is $-\frac{1}{27}$.