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A curve is defined by the parametric equations $$x=2t+\frac{1}{t^2}$$ $$y=2t-\frac{1}{t^2}$$ Show that the curve has the Cartesian equation $(x-y)(x+y)^2=k$

So I understand I need to eliminate the parameter $t$, but I'm not seeing an easy way to do this as I cannot rearrange for $t$ and then substitute. Any help will be appreciated.

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$$x-y=\frac{2}{t^2}$$ $$(x+y)^2=(4t)^2=16t^2$$ $$(x-y)(x+y)^2=\frac{2}{t^2}(16t^2)=32$$

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Whenever is given a parametric system $$\begin{cases}x=a(t)+b(t)\\y=a(t)-b(t),\;\end{cases}$$ computing $x+y\;$ and $\;x-y\;$ can help to eliminate the parameter.

In the present case $$\begin{cases}x+y=4t\\x-y={2\over{t^2}}\end{cases}$$ Since $(x-y)=k(x+y)^{-2},$ we obtain the constant as $(x-y)(x+y)^2.$

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