Step 1: Understanding Bijectivity

A function is bijective if it is both injective (one-to-one) and surjective (onto).

Step 2: Proving Injective (One-to-One)

Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$. Then:

Taking the cube root on both sides, we get $x_1 = x_2$. Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function is injective.

Step 3: Proving Surjective (Onto)

For any $y \in R$ (codomain), we need to find an $x \in R$ (domain) such that $f(x) = y$.

Since the cube root of any real number is a real number, for every $y \in R$, there exists $x = y^{1/3} \in R$ such that $f(x) = (y^{1/3})^3 = y$. Thus, the function is surjective.

Step 4: Conclusion

Since the function is both injective and surjective, it is bijective.