If $$3a-\frac{3}{a}-3=0$$, then what is the value of $$a^3-\frac{1}{a^3}+2$$ ?
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)
Create a FREE account and get:
