A curve is defined by the parametric equations $x=2t+\frac{1}{t^2},\; y=2t-\frac{1}{t^2}$. Find the Cartesian equation.

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.

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