if x^2+1/2x=cos theta then x^6+1/2x^3=

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Please find below the solution to the asked query:

x2+12x=cosθ...iiCube both sidesx2+12x3=cos3θAs a+b3=a3+b3+3a2b+3ab2x6+18x3+3x4.12x+3x2.14x2=cos3θx6+18x3+3x32+34=cos3θMultiply both sides 44x6+12x3+6x3+3=4cos3θ....iiWe know thatcos3θ=4cos3θ-3cosθ=4x6+12x3+6x3+3-3x2+12x Use i and ii=4x6+12x3+6x3-3x2+3-32x=x6+12x3+3x6+6x3-3x2+3-32xcos3θ=x6+12x3+3x6+2x3-x2+1-12xx6+12x3=cos3θ-3x6+2x3-x2+1-12x=cos3θ-3x6+2x3+1-x2+12x=cos3θ-3x6+2x3+1-cosθ=cos3θ+3cosθ-3x6+2x3+1x6+12x3=cos3θ+3cosθ-3x3+12

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