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]