If $$3x - \frac{1}{3x} = 9$$, then what is the value of $$x^{2} + \frac{x^2}{81}$$?
Given : $$3x-\frac{1}{3x}=9$$
Dividing both sides by 3, => $$x-\frac{1}{9x}=3$$
Squaring both sides, we get :
=>Â $$(x-\frac{1}{9x})^2=(3)^2$$
=> $$x^2+\frac{1}{81x^2}-2(x)(\frac{1}{9x})=9$$
=> $$(x^2+\frac{1}{81x^2})-\frac{2}{9}=9$$
=> $$(x^2+\frac{1}{81x^2})=9+\frac{2}{9}$$
=> $$(x^2+\frac{1}{81x^2})=\frac{83}{9}$$
=> Ans - (B)
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