We know $f(x) = x^3$. Now, I'm trying to show that $\mathbb{B}_n(f) \to f$.
I've shown this: $$B_n(f; x) = x^3 + \frac{1}{n^2}x(1 - x)(1 + (3n - 2)x$$ So we have $$\left| B_n(f) - f \right| = \left|x^3 + \frac{1}{n^2}x(1 - x)(1 + (3n - 2)x - x^3\right| = \frac{1}{n^2} \left|x(1 - x)(1 + (3n - 2)x\right|$$ from here I do not know what can I do.
Any hints will be appreciated.