Find the first three non-zero terms in the Maclaurin series for $\text{e}^x \sin(x + \pi)$. 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 expression using a trigonometric identity.

$\sin(x + \pi) = -\sin x$
$\text{e}^x \sin(x + \pi) = -\text{e}^x \sin x$

2. Expand using the standard Maclaurin series for $\text{e}^x$ and $\sin x$.

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

3. Multiply the series to extract the first three non-zero terms.

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

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

Using the addition formula, $\sin(x + \pi) = -\sin x$.
Using the standard Maclaurin series:
$e^x = 1 + x + \frac{x^2}{2!} + \dots$
$-\sin x = - \left(x - \frac{x^3}{3!} + \dots\right) = -x + \frac{x^3}{6} - \dots$
Multiplying the series up to the $x^3$ term:
$e^x \sin(x + \pi) = \left(1 + x + \frac{x^2}{2}\right)\left(-x + \frac{x^3}{6}\right) + \dots$
$= -x + \frac{x^3}{6} - x^2 - \frac{x^3}{2} + \dots$
$= -x - x^2 - \frac{1}{3}x^3 + \dots$
The first three non-zero terms are $-x$, $-x^2$, and $-\frac{1}{3}x^3$.

2. Expand $ax(1+bx)^c$ in ascending powers of $x$ up to $x^3$.

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

3. Equate the coefficients from both expansions to form a system of equations.

Equating the first three terms of this expansion to the result from part (a):
Coefficient of $x$: $\quad a = -1$
Coefficient of $x^2$: $\quad abc = -1$
Coefficient of $x^3$: $\quad \frac{ab^2c(c-1)}{2} = -\frac{1}{3}$

4. Solve the system of equations to find the exact values of $a$, $b$, and $c$.

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

5. Find the coefficient of $x^4$ using the established values of $a$, $b$, and $c$.

Substitute the exact values into the expression:
$ax(1+bx)^c = -x\left(1+\frac{1}{3}x\right)^3$
The $x^4$ term arises from multiplying the outer factor $-x$ by the $x^3$ term inside the expansion of $\left(1+\frac{1}{3}x\right)^3$.
The $x^3$ term is:
$\binom{3}{3}\left(\frac{1}{3}x\right)^3 = \frac{1}{27}x^3$
Multiplying by $-x$ gives $-\frac{1}{27}x^4$.
Therefore, the coefficient of $x^4$ is $-\frac{1}{27}$.