Given : $$2x+\frac{2}{9x}=4$$
=> $$x+\frac{1}{9x}=\frac{4}{2}=2$$
Multiplying both sides by 3,
=> $$3x+\frac{1}{3x}=6$$ ------------(i)
Cubing both sides, we get :
=> $$(3x+\frac{1}{3x})^3=(6)^3$$
=> $$27x^3+\frac{1}{27x^3}+3(3x)(\frac{1}{3x})(3x+\frac{1}{3x})=216$$
=> $$27x^3+\frac{1}{27x^3}+3(6)=216$$ Â Â [Using (i)]
=> $$27x^3+\frac{1}{27x^3}=216-18=198$$
=> Ans - (B)
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