1. **Problem Statement:** Find the value of $\alpha^3 + \beta^3$ where $\alpha, \beta$ are roots of the quadratic equation $$3x^2 - 2x + 4 = 0.$$ 2. **Recall the sum and product of roots formulas:** For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, we have: $$\alpha + \beta = -\frac{b}{a}, \quad \alpha \beta = \frac{c}{a}.$$ 3. **Calculate sum and product of roots:** $$\alpha + \beta = -\frac{-2}{3} = \frac{2}{3}, \quad \alpha \beta = \frac{4}{3}.$$ 4. **Use the identity for sum of cubes:** $$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta).$$ 5. **Substitute values:** $$\alpha^3 + \beta^3 = \left(\frac{2}{3}\right)^3 - 3 \times \frac{4}{3} \times \frac{2}{3} = \frac{8}{27} - 3 \times \frac{8}{9} = \frac{8}{27} - \frac{24}{9}.$$ 6. **Simplify:** $$\frac{8}{27} - \frac{24}{9} = \frac{8}{27} - \frac{72}{27} = -\frac{64}{27}.$$ **Final answer:** $$\boxed{-\frac{64}{27}}.$$ --- 1. **Problem Statement:** Find the positive integral number whose sum of squares of digits is 65 and the number is 9 times the sum of its digits. 2. **Let the two-digit number be $10x + y$ where $x,y$ are digits.** 3. **Given conditions:** $$x^2 + y^2 = 65,$$ $$10x + y = 9(x + y).$$ 4. **Simplify the second equation:** $$10x + y = 9x + 9y \implies 10x - 9x = 9y - y \implies x = 8y.$$ 5. **Since $x,y$ are digits (0 to 9), $x=8y$ implies $y=1$ and $x=8$.** 6. **Check sum of squares:** $$8^2 + 1^2 = 64 + 1 = 65,$$ which satisfies the first condition. 7. **Check the number:** $$10 \times 8 + 1 = 81,$$ $$9(8 + 1) = 9 \times 9 = 81,$$ which satisfies the second condition. **Final answer:** $$\boxed{81}.$$ --- 1. **Problem Statement:** Solve the simultaneous equations: $$2x + y = 1,$$ $$x^2 + y^2 = 10.$$ 2. **From the first equation, express $y$ in terms of $x$:** $$y = 1 - 2x.$$ 3. **Substitute into the second equation:** $$x^2 + (1 - 2x)^2 = 10.$$ 4. **Expand and simplify:** $$x^2 + 1 - 4x + 4x^2 = 10,$$ $$5x^2 - 4x + 1 = 10,$$ $$5x^2 - 4x + 1 - 10 = 0,$$ $$5x^2 - 4x - 9 = 0.$$ 5. **Solve quadratic equation:** $$x = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 5 \times (-9)}}{2 \times 5} = \frac{4 \pm \sqrt{16 + 180}}{10} = \frac{4 \pm \sqrt{196}}{10} = \frac{4 \pm 14}{10}.$$ 6. **Calculate roots:** $$x_1 = \frac{4 + 14}{10} = \frac{18}{10} = 1.8,$$ $$x_2 = \frac{4 - 14}{10} = \frac{-10}{10} = -1.$$ 7. **Find corresponding $y$ values:** For $x=1.8$, $$y = 1 - 2(1.8) = 1 - 3.6 = -2.6.$$ For $x=-1$, $$y = 1 - 2(-1) = 1 + 2 = 3.$$ **Final solutions:** $$(x,y) = (1.8, -2.6) \text{ or } (-1, 3).$$ --- 1. **Problem Statement:** Evaluate $$\frac{1}{z - z^2}$$ when $$z = \frac{1 - i}{10}.$$ 2. **Calculate $z^2$:** $$z^2 = \left(\frac{1 - i}{10}\right)^2 = \frac{(1 - i)^2}{100} = \frac{1 - 2i + i^2}{100} = \frac{1 - 2i - 1}{100} = \frac{-2i}{100} = -\frac{i}{50}.$$ 3. **Calculate $z - z^2$:** $$z - z^2 = \frac{1 - i}{10} - \left(-\frac{i}{50}\right) = \frac{1 - i}{10} + \frac{i}{50} = \frac{5(1 - i)}{50} + \frac{i}{50} = \frac{5 - 5i + i}{50} = \frac{5 - 4i}{50}.$$ 4. **Simplify the expression:** $$\frac{1}{z - z^2} = \frac{1}{\frac{5 - 4i}{50}} = \frac{50}{5 - 4i}.$$ 5. **Rationalize the denominator:** $$\frac{50}{5 - 4i} \times \frac{5 + 4i}{5 + 4i} = \frac{50(5 + 4i)}{(5)^2 + (4)^2} = \frac{50(5 + 4i)}{25 + 16} = \frac{50(5 + 4i)}{41} = \frac{250 + 200i}{41}.$$ **Final answer:** $$\boxed{\frac{250}{41} + \frac{200}{41}i}.$$