1. **State the problem:** We are given a quartic equation $$x^4 - 2x^3 + x^2 - 4 = 0$$ with roots $a, \beta, \gamma, \delta$. We want to show that $$S_4 = 9S_3$$ where $$S_k = a^k + \beta^k + \gamma^k + \delta^k$$. 2. **Identify coefficients:** The polynomial is $$x^4 - 2x^3 + x^2 + 0x - 4 = 0$$. So the coefficients are: $$a_3 = -2, \quad a_2 = 1, \quad a_1 = 0, \quad a_0 = -4$$. 3. **Use Newton's identities:** For a quartic with roots $r_i$, the power sums satisfy: $$S_1 + a_3 = 0$$ $$S_2 + a_3 S_1 + 2 a_2 = 0$$ $$S_3 + a_3 S_2 + a_2 S_1 + 3 a_1 = 0$$ $$S_4 + a_3 S_3 + a_2 S_2 + a_1 S_1 + 4 a_0 = 0$$ 4. **Calculate $S_1$:** $$S_1 + (-2) = 0 \implies S_1 = 2$$ 5. **Calculate $S_2$:** $$S_2 + (-2)(2) + 2(1) = 0 \implies S_2 - 4 + 2 = 0 \implies S_2 - 2 = 0 \implies S_2 = 2$$ 6. **Calculate $S_3$:** $$S_3 + (-2)(2) + (1)(2) + 3(0) = 0 \implies S_3 - 4 + 2 = 0 \implies S_3 - 2 = 0 \implies S_3 = 2$$ 7. **Calculate $S_4$:** $$S_4 + (-2)(2) + (1)(2) + 0(2) + 4(-4) = 0$$ $$S_4 - 4 + 2 - 16 = 0 \implies S_4 - 18 = 0 \implies S_4 = 18$$ 8. **Verify the relation:** $$9 S_3 = 9 \times 2 = 18 = S_4$$ **Final answer:** We have shown that $$S_4 = 9 S_3$$ as required.