1. **State the problem:** Given that $\tan\theta + \cot\theta = 2$, find the value of $\tan^3\theta + \cot^3\theta$. 2. **Recall the formula:** The sum of cubes formula is: $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$ Here, let $a = \tan\theta$ and $b = \cot\theta$. 3. **Use the given information:** We know $a + b = 2$. 4. **Find $ab$:** Since $\cot\theta = \frac{1}{\tan\theta}$, then $$ab = \tan\theta \cdot \cot\theta = 1$$ 5. **Calculate $a^2 - ab + b^2$:** $$a^2 - ab + b^2 = (a + b)^2 - 3ab = 2^2 - 3 \times 1 = 4 - 3 = 1$$ 6. **Calculate $a^3 + b^3$:** $$a^3 + b^3 = (a + b)(a^2 - ab + b^2) = 2 \times 1 = 2$$ **Final answer:** $$\tan^3\theta + \cot^3\theta = 2$$