Factorise $x^4 - 1$ completely. Given that $\frac{x^4}{x^4 - 1} = a + \frac{b}{x^4 - 1}$, find the values of $a$ and of $b$. Express $\frac{x^4}{x^4 - 1}$ in partial fractions.

1. Apply the difference of two squares identity to $x^4 - 1$

$x^4 - 1 = (x^2)^2 - 1^2$
$= (x^2 - 1)(x^2 + 1)$

2. Factorise $x^2 - 1$ completely

$x^2 - 1 = (x - 1)(x + 1)$
$\therefore x^4 - 1 = (x - 1)(x + 1)(x^2 + 1)$

1. Rewrite the numerator to match the denominator for easy division

$\frac{x^4}{x^4 - 1} = \frac{x^4 - 1 + 1}{x^4 - 1}$

2. Separate the fraction and simplify to find $a$ and $b$

$\frac{x^4 - 1 + 1}{x^4 - 1} = \frac{x^4 - 1}{x^4 - 1} + \frac{1}{x^4 - 1} = 1 + \frac{1}{x^4 - 1}$
Comparing with $a + \frac{b}{x^4 - 1}$, we get $a = 1$ and $b = 1$.

1. Use the results from parts (i) and (ii) to set up the partial fraction decomposition

From (i), $x^4 - 1 = (x - 1)(x + 1)(x^2 + 1)$.
From (ii), $\frac{x^4}{x^4 - 1} = 1 + \frac{1}{x^4 - 1}$.
We decompose $\frac{1}{x^4 - 1}$:
$\frac{1}{(x - 1)(x + 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{Cx + D}{x^2 + 1}$

2. Form the basic equation to solve for constants

$1 = A(x + 1)(x^2 + 1) + B(x - 1)(x^2 + 1) + (Cx + D)(x - 1)(x + 1)$

3. Solve for A and B by substituting suitable values of x

Let $x = 1$:
$1 = A(2)(2) \implies 4A = 1 \implies A = \frac{1}{4}$

Let $x = -1$:
$1 = B(-2)(2) \implies -4B = 1 \implies B = -\frac{1}{4}$

4. Solve for C and D by comparing coefficients or substituting other values

Let $x = 0$:
$1 = A(1)(1) + B(-1)(1) + D(-1) 1 = \frac{1}{4} + \frac{1}{4} - D 1 = \frac{1}{2} - D \implies D = -\frac{1}{2}$

Compare coefficients of $x^3$:
$0 = A + B + C 0 = \frac{1}{4} - \frac{1}{4} + C \implies C = 0$

5. Write the final expression in partial fractions

$\frac{x^4}{x^4 - 1} = 1 + \frac{1}{4(x-1)} - \frac{1}{4(x+1)} - \frac{1}{2(x^2+1)}$