Expand $\dfrac{1 - x^2}{\sqrt[3]{1 + 3x}}$ in ascending powers of $x$ up to and including the term in $x^3$. (ii) State the set of values of $x$ for which the series expansion is valid. (iii) Write down the equation of the tangent at the point $(0,1)$ on the curve $y = \dfrac{1 - x^2}{\sqrt[3]{1 + 3x}}.$

1. Rewrite the expression with a negative fractional index to prepare for expansion.

$\frac{1 - x^2}{\sqrt[3]{1 + 3x}} = (1 - x^2)(1 + 3x)^{-\frac{1}{3}}$

2. Expand $(1+3x)^{-\frac{1}{3}}$ up to the term in $x^3$.

$(1 + 3x)^{-\frac{1}{3}} = 1 + \left(-\frac{1}{3}\right)(3x) + \frac{\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{2!}(3x)^2 + \frac{\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)\left(-\frac{7}{3}\right)}{3!}(3x)^3 + \dots$
$= 1 - x + \frac{4/9}{2}(9x^2) - \frac{28/27}{6}(27x^3) + \dots$
$= 1 - x + 2x^2 - \frac{14}{3}x^3 + \dots$

3. Multiply by $(1-x^2)$ and collect terms up to $x^3$.

$(1 - x^2)(1 + 3x)^{-\frac{1}{3}} = (1 - x^2)\left(1 - x + 2x^2 - \frac{14}{3}x^3 + \dots\right)$
$= \left(1 - x + 2x^2 - \frac{14}{3}x^3\right) - x^2(1 - x) + \dots$
$= 1 - x + 2x^2 - \frac{14}{3}x^3 - x^2 + x^3$
$= 1 - x + x^2 - \frac{11}{3}x^3$

4. Determine the set of values of $x$ for which the expansion is valid.

The expansion is valid when $|3x| < 1$.
$|x| < \frac{1}{3} \implies -\frac{1}{3} < x < \frac{1}{3}$

5. Find the gradient of the tangent at $x = 0$ using the series expansion.

Near $x = 0$, $y \approx 1 - x + x^2 - \frac{11}{3}x^3$.
By differentiating, $\frac{dy}{dx} \approx -1 + 2x - 11x^2$.
At $x = 0$, the gradient $m = -1$.

(Alternatively, the coefficient of $x$ in the Maclaurin expansion directly gives $f'(0) = -1$.)

6. Form the equation of the tangent at the point $(0,1)$.

$y - 1 = -1(x - 0)$
$y = 1 - x$