Show that the function $f : R \rightarrow R$ given by $f(x) = x^{3}$ is injective.
(N/A) To check if the function $f(x) = x^{3}$ is injective (one-one),we assume $f(x_1) = f(x_2)$ for some $x_1, x_2 \in R$.
$f(x_1) = f(x_2) \Rightarrow x_1^{3} = x_2^{3}$.
Taking the cube root on both sides,we get $x_1 = x_2$.
Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in R$,the function $f$ is injective.
$f(x_1) = f(x_2) \Rightarrow x_1^{3} = x_2^{3}$.
Taking the cube root on both sides,we get $x_1 = x_2$.
Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in R$,the function $f$ is injective.
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