Question

if
${x}^{2} + \frac{1}{ {x}^{2} } = 51 \: then \: find \: \\ {x}^{3} - \frac{1}{ {x}^{3} }$
answer as soon as possible please

Answer 1

Here, such equations are followed :-

$(a - b)^2 = a^2 - 2ab + b^2 \\ \\ a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

In this question, a = x and b = 1/x .

$x^2 + \frac{1}{x^2} = 51 \\ \\ = (x)^2 + (\frac{1}{x})^2 = 51 \\ \\ \\ (x)^2 + (\frac{1}{x})^2 - (2 \times x \times \frac{1}{x}) = 51 - (2 \times x \times \frac{1}{x}) \\ \\ = x^2 + \frac{1}{x^2} - 2 = 51 - 2 \\ \\ = (x - \frac{1}{x})^2 = 49 \\ \\ \\ x - \frac{1}{x} = \sqrt{49} \\ \\ = x - \frac{1}{x} = 7$

$(x)^2 + (\frac{1}{x})^2 + (x \times \frac{1}{x}) = 51 + (x \times \frac{1}{x}) \\ \\ = x^2 + \frac{1}{x^2} + 1 = 51 + 1 \\ \\ = x^2 + \frac{1}{x^2} + 1 = 52 \\ \\ \\ (x^2 + \frac{1}{x^2} + 1)(x - \frac{1}{x}) = 52 \times 7 \\ \\ = (x)^3 - (\frac{1}{x})^3 = 52 \times 7 \\ \\ = x^3 - \frac{1}{x^3} = 364$

364 is the answer.

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