If cosA+cosB+cosC=0 and cos3A+cos3B+cos3C=λcosAcosBcosC then λ is
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If cosA+cosB+cosC=0 then cos3A+cos3B+cos3C=3cosAcosBcosC Given cos3A+cos3B+cos3c=λcosAcosBcosC4cos3A−3cosA+4cos3B−3cosB+4cos3C−3cosC=λcosAcosBcosC=4cos3A+cos3B+cos3C−3(cosA+cosB+cosC)=4(3cosAcosBcosC)−3(0)=12cosAcosBcosC