verify?!

**i_zz_y_ill** · October 2nd 2008, 04:42 AM

I cannot verify this formula. Can't figure it out at all even though its seems simple

Q
$z=\sqrt[3]{(4+\sqrt15)} +\sqrt[3]{(4-\sqrt15)}$ satisfies

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

**Soroban** · October 2nd 2008, 08:06 AM

Hello, i_zz_y_ill!

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

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

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

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

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

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

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

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

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