Let  cos - 1 x + cos - 1 2 x + cos - 1 3 x = π . If x satisfies the cubic ax 3 + bx 2 + cx – 1 = 0, then find of a + b + c.

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cos-1x+cos-12x+cos-13x=πLet A=cos-1xcosA=xB=cos-12xcosB=2xC=cos-13xcosC=3xAlso, A+B+C=πA+B=π-Ccos(A+B)=cosπ-CcosA cosB-sinA sinB=cosπ-CcosA cosB-sinA sinB=-cosCcosA cosB- sin2A sin2B=-cosCcosA cosB-1-cos2A1-cos2B=-cosCx×2x-1-x21-2x2=-3x2x2-1-x21-4x2=-3x2x2+3x=1-x21-4x2Squaring both sides, we get2x2+3x2=1-x21-4x24x4+9x2+12x3=1-4x2-x2+4x412x3+14x2-1=0Compare it with ax3+bx2+cx-1=0we get, a=12,b=14,c=0So, a+b+c=12+14+0=26
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