Question 88

If $$3a-\frac{3}{a}-3=0$$, then what is the value of $$a^3-\frac{1}{a^3}+2$$ ?

Solution

Given : $$3a-\frac{3}{a}-3=0$$

=> $$3(a-\frac{1}{a})=3$$

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

Cubing both sides, we get :

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

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

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

Adding '2' on both sides,

=> $$a^3-\frac{1}{a^3}+2=1+3+2=6$$

=> Ans - (D)

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