JEE Main 2023 — Mathematics Algebra

Given that roots of ${x}^{3}+bx+c=0$x3+bx+c=0 are $\alpha ,\beta ,\gamma$α,β,γ

Also given that

$$\beta \gamma =1=-\alpha$$βγ=1=−α

$$\Rightarrow \alpha =-1...(i)$$⇒α=−1...(i)

Now let us apply the relation between the roots of the cubic equation with the coefficients.

$$\Rightarrow \alpha +\beta +\gamma =0...(\mathrm{ii})$$⇒α+β+γ=0...(ii)

$$\Rightarrow \alpha \beta \gamma =-c...(\mathrm{iii})$$⇒αβγ=−c...(iii)

$$\Rightarrow (-1)(1)=-c$$⇒(−1)(1)=−c

$$\therefore c=1...(\mathrm{iv})$$∴c=1...(iv)

On substituting the value of $\alpha =-1$α=−1 in eq $(ii)$(ii) we get,

$$\beta +\gamma =1...(v)$$β+γ=1...(v)

$$\because \alpha \beta +\beta \gamma +\gamma \alpha =b$$∵αβ+βγ+γα=b

$$\Rightarrow \alpha (\beta +\gamma )+\beta \gamma =b$$⇒α(β+γ)+βγ=b

$$\therefore b=0...(\mathrm{vi})$$∴b=0...(vi)

Hence, equation will be,

$${x}^{3}+1=0$$x3+1=0

Whose roots will be, $-1,-\omega ,-{\omega }^{2}$−1,−ω,−ω2

We know that $1+\omega +{\omega }^{2}=0$1+ω+ω2=0

$$\Rightarrow \alpha +\beta +\gamma =0$$⇒α+β+γ=0

$$\Rightarrow -1+\beta +\gamma =0$$⇒−1+β+γ=0

$$\therefore \beta =-\omega ,\gamma =-{\omega }^{2}...(\mathrm{vii})$$∴β=−ω,γ=−ω2...(vii)

$\Rightarrow {\beta }^{3}=-{\omega }^{3}=-1$⇒β3=−ω3=−1 and $\Rightarrow {\gamma }^{3}=-{\omega }^{6}=-1$⇒γ3=−ω6=−1

Hence the value of

${b}^{3}+2{c}^{3}-3{\alpha }^{3}-6{\beta }^{3}-8{\gamma }^{3}$b3+2c3−3α3−6β3−8γ3 is

$$=0+2+3+6+8$$=0+2+3+6+8

$$=19$$=19

Hence this is the correct option.