alternative proof - Prove that $lim_{x to 0} frac{x - sin x}{x^3} = frac16$ using notable limits

Can we prove that $\lim_{x \to 0} \frac{x - \sin x}{x^3} = \frac 16$ using just algebra, trigonometric theorems and notable limits $\left( \mathrm{i.e. }\quad \frac{\sin x}{x} \to 1 \quad \mathrm{and} \quad \frac{1 - \cos x}{x^2} \to \frac 12 \right)$ and no l'Hospital rule, no series expansion, no Taylor series?