If 3x + frac{3}{x} = 1,then x^{3} + frac{1}{x | vedclass

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(B) Given the equation: $3x + \frac{3}{x} = 1$.Divide the entire equation by $3$ to get: $x + \frac{1}{x} = \frac{1}{3}$.Now,cube both sides of the eq

Vedclass · Accepted Answer

$\frac{1}{27}$

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(B) Given the equation: $3x + \frac{3}{x} = 1$.Divide the entire equation by $3$ to get: $x + \frac{1}{x} = \frac{1}{3}$.Now,cube both sides of the equation using the identity $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$:$(x + \frac{1}{x})^{3} = (\frac{1}{3})^{3}$$x^{3} + \frac{1}{x^{3}} + 3(x)(\frac{1}{x})(x + \frac{1}{x}) = \frac{1}{27}$Substitute the value $x + \frac{1}{x} = \frac{1}{3}$ into the equation:$x^{3} + \frac{1}{x^{3}} + 3(1)(\frac{1}{3}) = \frac{1}{27}$$x^{3} + \frac{1}{x^{3}} + 1 = \frac{1}{27}$.

If $3x + \frac{3}{x} = 1$,then $x^{3} + \frac{1}{x^{3}} + 1$ is

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