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If I am given, $x + y = a$ and $xy = b$, how would I find the value of $\dfrac1{x^3} + \dfrac1{y^3}$?

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\begin{align} \dfrac1{x^3} + \dfrac1{y^3} & = \dfrac{x^3+y^3}{(xy)^3} = \dfrac{\left(x+y \right)\left(x^2+y^2-xy \right)}{(xy)^3}\\ & = \dfrac{\left(x+y \right)\left(\left(x+y \right)^2-3xy \right)}{(xy)^3} = \dfrac{a\left(a^2-3b \right)}{b^3} \end{align}

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