Hello, please consider the following.
We know that
[tex]a = p^{\frac{1}{3}}-p^{-\frac{1}{3}}\\\\=p^{\frac{1}{3}}-\dfrac{1}{p^{\frac{1}{3}}}[/tex]
And we can write that.
[tex](p-\dfrac{1}{p})^3=(p-\dfrac{1}{p})(p^2-2+\dfrac{1}{p^2})\\\\=p^3-2p+\dfrac{1}{p}-p+\dfrac{2}{p}-\dfrac{1}{p^3}\\\\=p^3-\dfrac{1}{p^3}-3(p-\dfrac{1}{p})[/tex]
It means that, by replacing p by [tex]p^{1/3}[/tex]
[tex](p^{1/3}-\dfrac{1}{p^{1/3}})^3=p-\dfrac{1}{p}-3(p^{1/3}-\dfrac{1}{p^{1/3}})\\\\\\\text{ So }\\\\a^3=p-\dfrac{1}{p}-3a\\\\<=>\boxed{ a^3+3a=p-\dfrac{1}{p} }[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
