Question

1 If a = p^1/3-p^-1/3prove that: a^3 + 3a = p - 1/p​

Answer 1

Hello, please consider the following.

We know that

$a = p^{\frac{1}{3}}-p^{-\frac{1}{3}}\\\\=p^{\frac{1}{3}}-\dfrac{1}{p^{\frac{1}{3}}}$

And we can write that.

$(p-\dfrac{1}{p})^3=(p-\dfrac{1}{p})(p^2-2+\dfrac{1}{p^2})\\\\=p^3-2p+\dfrac{1}{p}-p+\dfrac{2}{p}-\dfrac{1}{p^3}\\\\=p^3-\dfrac{1}{p^3}-3(p-\dfrac{1}{p})$

It means that, by replacing p by $p^{1/3}$

$(p^{1/3}-\dfrac{1}{p^{1/3}})^3=p-\dfrac{1}{p}-3(p^{1/3}-\dfrac{1}{p^{1/3}})\\\\\\\text{ So }\\\\a^3=p-\dfrac{1}{p}-3a\\\\\boxed{ a^3+3a=p-\dfrac{1}{p} }$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you