1. **Problem statement:** Solve the equation $$\sqrt[5]{(3x^2 + 3x - 1)^3} = \sqrt[5]{x^3}$$ and choose all valid solutions. 2. **Step 1: Understand the equation** The equation involves fifth roots on both sides. Since fifth root is a one-to-one function, we can equate the radicands (the expressions inside the roots) raised to the power of 3 and 3 respectively: $$\sqrt[5]{a} = \sqrt[5]{b} \implies a = b$$ Here, $$a = (3x^2 + 3x - 1)^3$$ and $$b = x^3$$. 3. **Step 2: Equate the radicands:** $$ (3x^2 + 3x - 1)^3 = x^3 $$ 4. **Step 3: Take cube root on both sides:** $$ \sqrt[3]{(3x^2 + 3x - 1)^3} = \sqrt[3]{x^3} $$ which simplifies to $$ 3x^2 + 3x - 1 = x $$ 5. **Step 4: Rearrange the equation:** $$ 3x^2 + 3x - 1 - x = 0 $$ $$ 3x^2 + 2x - 1 = 0 $$ 6. **Step 5: Solve the quadratic equation:** Use the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $$a=3$$, $$b=2$$, $$c=-1$$. Calculate discriminant: $$ \Delta = 2^2 - 4 \times 3 \times (-1) = 4 + 12 = 16 $$ So, $$ x = \frac{-2 \pm \sqrt{16}}{2 \times 3} = \frac{-2 \pm 4}{6} $$ 7. **Step 6: Find the two roots:** - For $$+$$ sign: $$ x = \frac{-2 + 4}{6} = \frac{2}{6} = \frac{1}{3} $$ - For $$-$$ sign: $$ x = \frac{-2 - 4}{6} = \frac{-6}{6} = -1 $$ 8. **Step 7: Check for extraneous solutions:** Since the original equation involves fifth roots, which are defined for all real numbers, and the fifth root is one-to-one, both solutions are valid. 9. **Final answer:** $$ x = -1 \quad \text{or} \quad x = \frac{1}{3} $$