Question 4

What is the value of $$x^{3}-\frac{1}{x^{3}}+8$$ when $$8x-\frac{8}{x}-16=0$$ ?

Solution

Given : $$8x-\frac{8}{x}-16=0$$

=> $$x-\frac{1}{x}=2$$ --------------(i)

Cubing both sides, we get :

=> $$(x-\frac{1}{x})^3=(2)^3$$

=> $$(x^3-\frac{1}{x^3})-3(x)(\frac{1}{x})(x-\frac{1}{x})=8$$

Substituting value from equation (i),

=> $$(x^3-\frac{1}{x^3})-3(1)(2)=8$$

=> $$x^3-\frac{1}{x^3}=8+6=14$$

$$\therefore$$ $$x^{3}-\frac{1}{x^{3}}+8=22$$

=> Ans - (D)

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