# If x square+1/4x square=8.Find the value of x cube +1/8x cubeâ€‹

$\\ x^2 + \frac{1}{4x^2} = 8\\ (x)^2 + (1/2x)^2 + 1 = 9\ \ \ (\text{adding 1 on both sides})\\ \left( x + \frac{1}{2x}\right)^2 = 9\\ x + \frac{1}{2x} = 3 \\ \text{ cubing both sides}\\ (x)^3 + \left( \frac{1}{2x}\right)^3 =3^3\\ (x)^3 + \left( \frac{1}{2x}\right)^3 + 3(x)\frac{1}{2x}(x + \frac{1}{2x}) = 27\\ x^3 + \frac{1}{8x^3} + \frac{3}{2}\times 3 = 27\\ x^3 + \frac{1}{8x^3} = 27 - \frac{9}{2} =\frac{45}{2}$

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