JEE Main 20238 Apr, Morning shiftMathematicsMediumComplex Numbers and Quadratic Equations264× since 2002Let α,β,γ\alpha, \beta, \gammaα,β,γ be the three roots of the equation x3+bx+c=0x^{3}+b x+c=0x3+bx+c=0. If βγ=1=−α\beta \gamma=1=-\alphaβγ=1=−α, then b3+2c3−3α3−6β3−8γ3b^{3}+2 c^{3}-3 \alpha^{3}-6 \beta^{3}-8 \gamma^{3}b3+2c3−3α3−6β3−8γ3 is equal to :AOption A. 21BOption B. 19COption C. 1698\frac{169}{8}8169DOption D. 1558\frac{155}{8}8155CheckSkip · show solution
