1. **Problem:** Find an equation whose roots are $a^3\beta$ and $\alpha\beta^3$, where $a$ and $\beta$ are roots of $2x^2 - 4x + 1 = 0$. 2. **Step 1: Identify sum and product of roots of the original equation.** The quadratic is $2x^2 - 4x + 1 = 0$. Sum of roots $a + \beta = -\frac{-4}{2} = 2$. Product of roots $a\beta = \frac{1}{2}$. 3. **Step 2: Find sum and product of new roots $a^3\beta$ and $a\beta^3$.** Sum: $$a^3\beta + a\beta^3 = a\beta(a^2 + \beta^2)$$ Recall $a^2 + \beta^2 = (a + \beta)^2 - 2a\beta = 2^2 - 2 \times \frac{1}{2} = 4 - 1 = 3$$ So sum = $a\beta \times 3 = \frac{1}{2} \times 3 = \frac{3}{2}$. Product: $$a^3\beta \times a\beta^3 = a^4 \beta^4 = (a\beta)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$$ 4. **Step 3: Form the quadratic equation with roots $a^3\beta$ and $a\beta^3$.** Using sum and product of roots, the quadratic is: $$x^2 - \left(\text{sum}\right)x + \text{product} = 0$$ $$x^2 - \frac{3}{2}x + \frac{1}{16} = 0$$ 5. **Step 4: Clear denominators for a nicer form.** Multiply entire equation by 16: $$16x^2 - 24x + 1 = 0$$ **Final answer:** $$16x^2 - 24x + 1 = 0$$ This is the quadratic equation whose roots are $a^3\beta$ and $a\beta^3$.