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Start learning 50% faster. Sign in nowx+ 1/x = 2cosθ (x+ 1/x)³ = (2cosθ)³ x³+ 1/x³+ 3x × 1/x ×(x+ 1/x) = 8cos³θ x³+ 1/x³+ 3 ×2cosθ = 8cos³θ x³+ 1/x³ = 8cos³θ - 6cosθ x³+ 1/x³ = 2(4cos³θ-3cosθ) Using identity [4cos³θ-3cosθ=cos3θ] x³+ 1/x³ = 2cos3θ
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Indian Institute of Sugarcane Research is located in?
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