Show that $f$ is a bijection.

Hence write down the inverse function ${f^{ - 1}}(x,{\text{ }}y)$.

for $f$ to be a bijection it must be both an injection and a surjection     R1

Note:     Award this R1 for stating this anywhere.

suppose that $f(a,{\text{ }}b) = f(c,{\text{ }}d)$     (M1)

it follows that

$2{a^3} + {b^3} = 2{c^3} + {d^3}$ and ${a^3} + 2{b^3} = {c^3} + 2{d^3}$     A1

attempting to solve the two equations     M1

to obtain $3{a^3} = 3{c^3}$

Note:     Award M1 only if a good attempt is made to solve the system.

$\Rightarrow a = c$ and therefore $b = d$     A1

$f$ is an injection because $f(a,{\text{ }}b) = f(c,{\text{ }}d) \Rightarrow (a,{\text{ }}b) = (c,{\text{ }}d)$     R1

Note:     Award this R1 for stating this anywhere providing that an attempt is made to prove injectivity.

let $(p,{\text{ }}q) \in \mathbb{R} \times \mathbb{R}$ and let $f(r,{\text{ }}s) = (p,{\text{ }}q)$     (M1)

then, $p = 2{r^3} + {s^3}$ and $q = {r^3} + 2{s^3}$     A1

attempting to solve the two equations     M1

$r = \sqrt[3]{{\frac{{2p - q}}{3}}}$ and $s = \sqrt[3]{{\frac{{2q - p}}{3}}}$     A1A1

$f$ is a surjection because given $(p,{\text{ }}q) \in \mathbb{R} \times \mathbb{R}$, there exists $(r,{\text{ }}s) \in \mathbb{R} \times \mathbb{R}$ such that $f(r,{\text{ }}s) = (p,{\text{ }}q)$     R1

Note:   Award this R1 for stating this anywhere providing that an attempt is made to prove surjectivity.

[12 marks]

$\left( {{f^{ - 1}}(x,{\text{ }}y) = } \right)\,\,\,\left( {\sqrt[3]{{\frac{{2x - y}}{3},}}{\text{ }}\sqrt[3]{{\frac{{2y - x}}{3}}}} \right)$     A1

Note:     A1 for correct expressions in $x$ and $y$, allow FT only if the expression is deduced in part (a).

[1 mark]