pretty simple question
Take cube both sides,
therefore
8(cosA)^3=x^3+3(x+1/x)+1/x^3
8(cosA)^3=x^3+3(2cosA)+1/x^3
8(cosA)^3 - 3(2cosA)=x^3+1/x^3
2{4(cosA)^3 - 3(cosA)}=x^3+1/x^3
2cos3A=x^3+1/x^3....
Answer.:-2cos3A.. Hope this was the answer expected...