1. **Stating the problem:** Simplify the expression $$- x^{\frac{2}{3}} + x^{\frac{1}{3}}$$ and identify which of the given options it equals. 2. **Recall the properties of exponents:** - $x^{a} \cdot x^{b} = x^{a+b}$ - $x^{\frac{1}{3}}$ means the cube root of $x$, i.e., $\sqrt[3]{x}$ 3. **Rewrite the expression:** $$- x^{\frac{2}{3}} + x^{\frac{1}{3}} = - (x^{\frac{1}{3}})^2 + x^{\frac{1}{3}}$$ 4. **Let $y = x^{\frac{1}{3}}$ (cube root of $x$), then the expression becomes:** $$- y^2 + y = y - y^2$$ 5. **This expression is not equal to any single power of $x$ directly, but we can factor it:** $$y - y^2 = y(1 - y) = x^{\frac{1}{3}} (1 - x^{\frac{1}{3}})$$ 6. **Among the options:** - a. $x$ - b. $x^2$ - c. $x^3$ - d. $x^{\frac{3}{2}}$ - e. $\sqrt[3]{x}$ The expression is closest to $x^{\frac{1}{3}}$ but with an additional factor $(1 - x^{\frac{1}{3}})$, so it does not equal any single option exactly. **Final answer:** None of the options exactly equal the expression $- x^{\frac{2}{3}} + x^{\frac{1}{3}}$. --- **Example:** If $x=8$, then: - $x^{\frac{1}{3}} = 2$ - $x^{\frac{2}{3}} = (x^{\frac{1}{3}})^2 = 2^2 = 4$ So, $$- x^{\frac{2}{3}} + x^{\frac{1}{3}} = -4 + 2 = -2$$ This is not equal to any of the options evaluated at $x=8$.