\(\displaystyle{f}:{A}\to{B}\) is one-to-one if \(\displaystyle∀{a},{b}\in{A},{f{{\left({a}\right)}}}={f{{\left({b}\right)}}}\to{a}={b}\)

Assume that \(\displaystyle∀{x},{y}\in\mathbb{R},{f{{\left({x}\right)}}}={f{{\left({y}\right)}}}.\)

f(x)=f(y)

\(\displaystyle{x}^{{3}}={y}^{{3}}\)

x=y

Thus,

\(\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}\) is one-to-one

"For the function to be onto:

f(x)=y

let \(\displaystyle{y}\in\mathbb{R}\)

Choose x to be \(\displaystyle{y}^{{\frac{{1}}{{3}}}}.\)

Then,

\(\displaystyle{x}\in\mathbb{R}\)

and

\(\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}\)

\(\displaystyle={\left({y}^{{\frac{{1}}{{3}}}}\right)}^{{3}}\)

=y

Thus \(\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}\) is onto

Since it's both one-to-one and onto, it's a bijection function from \(\displaystyle\mathbb{R}\) to \(\displaystyle\mathbb{R}\)."

Assume that \(\displaystyle∀{x},{y}\in\mathbb{R},{f{{\left({x}\right)}}}={f{{\left({y}\right)}}}.\)

f(x)=f(y)

\(\displaystyle{x}^{{3}}={y}^{{3}}\)

x=y

Thus,

\(\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}\) is one-to-one

"For the function to be onto:

f(x)=y

let \(\displaystyle{y}\in\mathbb{R}\)

Choose x to be \(\displaystyle{y}^{{\frac{{1}}{{3}}}}.\)

Then,

\(\displaystyle{x}\in\mathbb{R}\)

and

\(\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}\)

\(\displaystyle={\left({y}^{{\frac{{1}}{{3}}}}\right)}^{{3}}\)

=y

Thus \(\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}\) is onto

Since it's both one-to-one and onto, it's a bijection function from \(\displaystyle\mathbb{R}\) to \(\displaystyle\mathbb{R}\)."