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

The cubic is monic, so if its roots are $$\alpha=1$$ and $$\beta = 1+i\sqrt{2}$$, then because $$a, b, c \in \mathbb{R}$$ the conjugate $$\overline{\beta}=1-i\sqrt{2}$$ must also be a root.

Thus the three roots are
$$r_1 = 1,\; r_2 = 1+i\sqrt{2},\; r_3 = 1-i\sqrt{2}.$$ The polynomial can be written as

$$x^3+ax^2+bx+c=(x-r_1)(x-r_2)(x-r_3).$$

First multiply the complex pair:

$$(x-r_2)(x-r_3)=\bigl(x-(1+i\sqrt{2})\bigr)\bigl(x-(1-i\sqrt{2})\bigr).$$

Set $$t=x-1$$ so that the product becomes $$(t-i\sqrt{2})(t+i\sqrt{2})=t^2+2.$$ Since $$t=x-1,$$ we have

$$t^2+2=(x-1)^2+2 =x^2-2x+1+2 =x^2-2x+3.$$

Now include the remaining factor:

$$(x-1)(x^2-2x+3) =x^3-2x^2+3x-x^2+2x-3 =x^3-3x^2+5x-3.$$

Hence $$a=-3,\; b=5,\; c=-3.$$

We now evaluate the definite integral

$$I=\int_{-1}^{1}\bigl(x^3+ax^2+bx+c\bigr)\,dx.$$

Integrate term by term:

$$\int_{-1}^{1}x^3\,dx =\left.\frac{x^4}{4}\right|_{-1}^{1} =\frac{1}{4}-\frac{1}{4}=0,$$

$$\int_{-1}^{1}ax^2\,dx =a\left.\frac{x^3}{3}\right|_{-1}^{1} =a\left(\frac{1}{3}-\frac{-1}{3}\right) =\frac{2a}{3},$$

$$\int_{-1}^{1}bx\,dx =b\left.\frac{x^2}{2}\right|_{-1}^{1} =b\left(\frac{1}{2}-\frac{1}{2}\right)=0,$$

$$\int_{-1}^{1}c\,dx =c\left.x\right|_{-1}^{1} =c(1-(-1))=2c.$$

Therefore

$$I=\frac{2a}{3}+2c =2\left(\frac{a}{3}+c\right).$$

Substituting $$a=-3$$ and $$c=-3$$ gives

$$I=2\left(\frac{-3}{3}+(-3)\right) =2(-1-3) =2(-4) =-8.$$

Hence the value of the integral is $$-8$$.

Option C which is: $$-8$$