What is the value of $$x^{3}-\frac{1}{x^{3}}+8$$ when $$8x-\frac{8}{x}-16=0$$ ?
Given : $$8x-\frac{8}{x}-16=0$$
=>Â $$x-\frac{1}{x}=2$$ --------------(i)
Cubing both sides, we get :
=>Â $$(x-\frac{1}{x})^3=(2)^3$$
=> $$(x^3-\frac{1}{x^3})-3(x)(\frac{1}{x})(x-\frac{1}{x})=8$$
Substituting value from equation (i),
=> $$(x^3-\frac{1}{x^3})-3(1)(2)=8$$
=> $$x^3-\frac{1}{x^3}=8+6=14$$
$$\therefore$$Â $$x^{3}-\frac{1}{x^{3}}+8=22$$
=> Ans - (D)
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