Σa3cos⁡(B−C)=

Σa3cos(BC)=

  1. A

     3 abc

  2. B

    3 (a+b+c) 

  3. C

    abc (a+ b + c)

  4. D

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    Solution:

    We have,

    a3cos(BC)=k3sin3Acos(BC)=k3Σsin2Asin(B+C)cos(BC)=k32Σsin2A(sin2B+sin2C)=k32Σsin2A(sin2B+sin2C)+sin2B(sin2C+sin2A)+sin2C(sin2A+sin2B)

    =k3Σsin2AsinBcosB+sin2BsinAcosA=k3ΣsinAsinBsin(A+B)=k3[sinAsinBsinC+sinBsinCsinA+sinCsinAsinB]=3(ksinA)(ksinB)(ksinC)=3abc

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