If alpha = 1 and beta = 1 + isqrt{2}, where i | vedclass

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(C) Since the coefficients $a, b, c$ are real, complex roots must occur in conjugate pairs. Given $\beta = 1 + i\sqrt{2}$ is a root, its conjugate $\b

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-$8$

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(C) Since the coefficients $a, b, c$ are real, complex roots must occur in conjugate pairs. Given $\beta = 1 + i\sqrt{2}$ is a root, its conjugate $\bar{\beta} = 1 - i\sqrt{2}$ must also be a root.The roots of the cubic equation are $1, 1 + i\sqrt{2}$, and $1 - i\sqrt{2}$.The polynomial can be written as $(x - 1)(x - (1 + i\sqrt{2}))(x - (1 - i\sqrt{2}))$.$= (x - 1)((x - 1)^2 - (i\sqrt{2})^2) = (x - 1)((x - 1)^2 + 2) = (x - 1)(x^2 - 2x + 1 + 2) = (x - 1)(x^2 - 2x + 3)$.$= x^3 - 2x^2 + 3x - x^2 + 2x - 3 = x^3 - 3x^2 + 5x - 3$.Comparing this with $x^3 + ax^2 + bx + c = 0$, we get $a = -3, b = 5, c = -3$.We need to evaluate $\int_{-1}^{1} (x^3 - 3x^2 + 5x - 3) dx$.Since $x^3$ and $5x$ are odd functions, their integral over the symmetric interval $[-1, 1]$ is $0$.Thus, the integral becomes $\int_{-1}^{1} (-3x^2 - 3) dx = 2 \int_{0}^{1} (-3x^2 - 3) dx$.$= 2 [-x^3 - 3x]_0^1 = 2(-1 - 3) = 2(-4) = -8$.

If $\alpha = 1$ and $\beta = 1 + i\sqrt{2}$, where $i = \sqrt{-1}$ are two roots of the equation $x^3 + ax^2 + bx + c = 0$ with $a, b, c \in R$, then $\int_{-1}^{1} (x^3 + ax^2 + bx + c) dx$ is equal to:

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