1. **State the problem:** Calculate $\sin^3 70^\circ + \cos^3 70^\circ$. 2. **Use the sum of cubes formula:** Recall that $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. Here, let $a = \sin 70^\circ$ and $b = \cos 70^\circ$. 3. **Apply the formula:** $$\sin^3 70^\circ + \cos^3 70^\circ = (\sin 70^\circ + \cos 70^\circ)(\sin^2 70^\circ - \sin 70^\circ \cos 70^\circ + \cos^2 70^\circ)$$ 4. **Simplify using Pythagorean identity:** Since $\sin^2 x + \cos^2 x = 1$, we have $$\sin^2 70^\circ + \cos^2 70^\circ = 1$$ So the expression becomes $$(\sin 70^\circ + \cos 70^\circ)(1 - \sin 70^\circ \cos 70^\circ)$$ 5. **Calculate $\sin 70^\circ + \cos 70^\circ$:** Using approximate values, $\sin 70^\circ \approx 0.9397$, $\cos 70^\circ \approx 0.3420$ So, $$0.9397 + 0.3420 = 1.2817$$ 6. **Calculate $\sin 70^\circ \cos 70^\circ$:** $$0.9397 \times 0.3420 = 0.3214$$ 7. **Substitute back:** $$(1.2817)(1 - 0.3214) = 1.2817 \times 0.6786 = 0.8697$$ 8. **Final answer:** $$\sin^3 70^\circ + \cos^3 70^\circ \approx 0.87$$