Let cos−1x+cos−12x+cos−13x=π. If x satisfies the equation ax3+bx2+c=0 then the value of (a + b + c) is
cos−1(x)+cos−1(2x)+cos−1(3x)=π or cos−1(2x)+cos−1(3x)=π−cos−1(x)=cos−1(−x) or cos−1(2x)(3x)−1−4x21−9x2=cos−1(−x) or 6x2−1−4x21−9x2=−x or 6x2+x2=1−4x21−9x2 or x2+12x3=1−13x2 or 12x3+14x2−1=0⇒ a=12;b=14;c=-1⇒ a+b+c=12+14−1=25