1. ## verify?!

I cannot verify this formula. Can't figure it out at all even though its seems simple Q
$\displaystyle z=\sqrt[3]{(4+\sqrt15)} +\sqrt[3]{(4-\sqrt15)}$ satisfies

$\displaystyle {z}^3-3z-8=0$

2. Hello, i_zz_y_ill!

Show that: .$\displaystyle z\:=\:\sqrt[3]{(4+\sqrt15)} +\sqrt[3]{(4-\sqrt15)}$ .satisfies .$\displaystyle z^3-3z-8\:=\:0$

Let: .$\displaystyle \begin{array}{ccccccc} u \;=\; \sqrt[3]{4 + \sqrt{15}} & \Rightarrow & u^3 \;=\; 4 + \sqrt{15} \\ \\[-4mm] v \;=\; \sqrt[3]{4-\sqrt{15}} & \Rightarrow & v^3 \;=\; 4 - \sqrt{15} \end{array}$ . . . Hence: .$\displaystyle z \:=\:u+v$

Then: .$\displaystyle u^3 + v^3 \:=\:8$

. and: .$\displaystyle uv \:=\:\sqrt[3]{(4 +\sqrt{15})(4-\sqrt{15})} \;=\;\sqrt[3]{16-15} \;=\;\sqrt[3]{1} \;=\;1$

We have: .$\displaystyle z^3 - 3z - 8 \;=\;(u+v)^3 - 3(u+v) - 8$

Expand: .$\displaystyle u^3 + 3u^2v + 3uv^2 - 3(u+v) -8$

$\displaystyle \text{Re-arrange terms: }\;\underbrace{(u^3+v^3)}_{\text{This is 8}} \;+ \;3\underbrace{(uv)}_{\text{This is 1}}(u+v) \;- \;3(u+v) \;-\;8$

. . . . . . . . . . $\displaystyle =\;\;8 + 3(u+v) - 3(u+v) -8\;\;=\;\;\boxed{0} \quad\hdots\;\text{There!}$