Question 65

If $$2x+\frac{2}{9x}=4$$, then the value of $$27x^3+\frac{1}{27x^3}$$ is

Solution

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 :

=> $$(3x+\frac{1}{3x})^3=(6)^3$$

=> $$27x^3+\frac{1}{27x^3}+3(3x)(\frac{1}{3x})(3x+\frac{1}{3x})=216$$

=> $$27x^3+\frac{1}{27x^3}+3(6)=216$$ [Using (i)]

=> $$27x^3+\frac{1}{27x^3}=216-18=198$$

=> Ans - (B)

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