Verified Solution • 2025–26 Session moderate

QJEEM23Ma0ecf: JEE Main 2023 Mathematics - Class 11 JEEM23Ma0ecf Solution: Let be the three roots of the equation . If , then is equal to :

Let α,β,γ\alpha, \beta, \gamma be the three roots of the equation x3+bx+c=0x^{3}+b x+c=0 . If βγ=1=−α\beta \gamma=1=-\alpha , then b3+2c3−3α3−6β3−8γ3b^{3}+2 c^{3}-3 \alpha^{3}-6 \beta^{3}-8 \gamma^{3} is equal to :

Multiple Choice Options

  • A 21
  • B 19
  • C 1698\frac{169}{8}
  • D 1558\frac{155}{8}

Step-by-Step Verified Solution

Given cubic equation is :

x3+bx+c=0x^3+b x+c=0

∵α,β,γ\because \alpha, \beta, \gamma are the roots of above equation.

And βγ=1=−α\beta \gamma=1=-\alpha

amp; So, product of roots =−camp;⇒αβγ=−camp;⇒(−1)(1)=−camp;⇒c=1\begin{aligned}& \text { So, product of roots }=-c \\\\& \Rightarrow \alpha \beta \gamma=-c \\\\& \Rightarrow(-1)(1)=-c \\\\& \Rightarrow c=1\end{aligned}

Since, α=−1\alpha=-1 is the root. So,

amp;⇒−1−b+c=0amp;⇒c−b=1amp;⇒1−b=1⇒b=0\begin{aligned}& \Rightarrow-1-b+c=0 \\\\& \Rightarrow c-b=1 \\\\& \Rightarrow 1-b=1 \Rightarrow b=0\end{aligned}

The given equation becomes x3+1=0x^3+1=0

So, roots are −1,−ω,−ω2-1,-\omega,-\omega^2

amp;∴b3+2c3−3α3−6β3−8γ3amp;=0+2−3(−1)3−6(−ω)3−8(−ω2)3amp;=2+3+6ω3+8ω6amp;=5+6+8=19\begin{aligned}& \therefore b^3+2 c^3-3 \alpha^3-6 \beta^3-8 \gamma^3 \\\\& =0+2-3(-1)^3-6(-\omega)^3-8\left(-\omega^2\right)^3 \\\\& =2+3+6 \omega^3+8 \omega^6 \\\\& =5+6+8=19\end{aligned}

Concept :

For a cubic equation, ax3+bx2+cx+d=0a x^3+b x^2+c x+d=0

Sum of roots =−ba=\frac{-b}{a}

Product of roots taken two at a time =ca=\frac{c}{a}

Product of roots =−da=\frac{-d}{a}

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