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David Cohen
suppose that
x+1/x=2cos(θ)
show that x^3+1/x^3=2cos(3θ)
Given,
x+1/x=2cos(θ)
=>(x+1/x)^3=(2cos(θ) )^3
(x+1/x)^3=(2cos(θ) )^3
x^3 + 1/x^3 + 3 x + 3/x=8cos^3(θ)
x^3 + 1/x^3 + 3(x + 1/x)=8cos^3(θ) ......substitute given value for (x+1/x)
x^3 + 1/x^3 + 3*2cos(θ)=8cos^3(θ)
x^3 + 1/x^3 + 6cos(θ)=8cos^3(θ)
x^3 + 1/x^3 =8cos^3(θ)-6cos(θ)
x^3 + 1/x^3 =2cos(theta)(4cos^2(θ)-3)
x^3 + 1/x^3 =2cos(θ)(2cos(2 θ) - 1)
x^3 + 1/x^3 =2cos3θ.