If frac{x}{y} + frac{y}{x} = 6, then find the | vedclass

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(B) Given that $\frac{x}{y} + \frac{y}{x} = 6.$We know the algebraic identity $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b).$Let $a = \frac{x}{y}$ and $b

Vedclass · Accepted Answer

$198$

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(B) Given that $\frac{x}{y} + \frac{y}{x} = 6.$We know the algebraic identity $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b).$Let $a = \frac{x}{y}$ and $b = \frac{y}{x}.$Then,$\left(\frac{x}{y} + \frac{y}{x}\right)^{3} = \frac{x^{3}}{y^{3}} + \frac{y^{3}}{x^{3}} + 3\left(\frac{x}{y}\right)\left(\frac{y}{x}\right)\left(\frac{x}{y} + \frac{y}{x}\right).$Since $\left(\frac{x}{y}\right)\left(\frac{y}{x}\right) = 1,$ the equation becomes:$(6)^{3} = \frac{x^{3}}{y^{3}} + \frac{y^{3}}{x^{3}} + 3(1)(6).$$216 = \frac{x^{3}}{y^{3}} + \frac{y^{3}}{x^{3}} + 18.$$\frac{x^{3}}{y^{3}} + \frac{y^{3}}{x^{3}} = 216 - 18 = 198.$

If $\frac{x}{y} + \frac{y}{x} = 6,$ then find the value of $\frac{x^{3}}{y^{3}} + \frac{y^{3}}{x^{3}}.$

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