If $\alpha =1$ and $\beta =1+i\sqrt{2}$, where $i=\sqrt{-1}$ are two roots of the equation ${x}^{3}+a{x}^{2}+bx+c=0,a,b,c\in \mathrm{R}$, then ${\int }_{-1}^{1}\left({x}^{3}+a{x}^{2}+bx+c\right)dx$ is equal to: [JEE Main 2026, 4 Apr (Shift 2)]

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YESPYQ · Accepted Answer

Correct answer: c) $–8$. Roots are $1 \pm \mathrm{i} \sqrt{2} $ and $1$ Sum of roots: $-\mathrm{a}=3 \Rightarrow \mathrm{a}=-3$ Product of roots: $-\mathrm{c}=1(1+\sqrt{2} \mathrm{i})(1-\mathrm{i} \sqrt{2})$ $\Rightarrow \mathrm{c}=-3$ $I=\int_{-1}^1\left(x^3-3 x^2+b x-3\right) d x$ $=2 \int_0^1\left(-3 x^2-3\right) d x$ $=2\left(-x^3-3 x\right)_0^1=-8$

If \(\alpha =1\) and \(\beta =1+i\sqrt{2}\), where \(i=\sqrt{-1}\) are two roots of the equation \({x}^{3}+a{x}^{2}+bx+c…

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