BigBeachBanana
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[math]\large \left(\sqrt[3]{4-\sqrt{15}}\right)^x + \left(\sqrt[3]{4+\sqrt{15}}\right)^x=8[/math]
Solve for [imath]\large x.[/imath]
Solve for [imath]\large x.[/imath]
Oops, my cube root signs need to be a bit longer.\(\displaystyle 1 + (\sqrt[3] 4+ \sqrt 15)^{2x} =8(\sqrt[3] 4+ \sqrt 15)^x\)
1+u^2 = 8u
\(\displaystyle u = 4 \pm \sqrt 15\)
\(\displaystyle (\sqrt[3] 4+ \sqrt 15)^x =4 \pm \sqrt 15\)
x=3 or \(\displaystyle \dfrac {3\ln(4-\sqrt 15)}{\ln(4+\sqrt 15)}\)
Very elegant derivation.....We observe that [math]\sqrt[3]{4-\sqrt{15}}\cdot\sqrt[3]{4+\sqrt{15}}=1\Rightarrow\sqrt[3]{4+\sqrt{15}}=\frac{1}{\sqrt[3]{4-\sqrt{15}}}[/math]Let [math]\left(\sqrt[3]{4-\sqrt{15}}\right)^x=t[/math]Then [math] t+\frac{1}{t}=8\Rightarrow t^2-8t+1=0\Rightarrow t_{1,2}=4\pm\sqrt{15}[/math][math]\left(\sqrt[3]{4-\sqrt{15}}\right)^x=4-\sqrt{15}\Rightarrow x=3[/math][math]\left(\sqrt[3]{4-\sqrt{15}}\right)^x=4+\sqrt{15}=\left(4-\sqrt{15}\right)^{-1}\Rightarrow x=-3[/math]
[math]\sqrt[3]{4-\sqrt{15}}\cdot\sqrt[3]{4+\sqrt{15}}[/math][math]=\sqrt[3]{\left(4-\sqrt{15}\right) \left(4+\sqrt{15}\right)}[/math][math]=\sqrt[3]{4^2-(\sqrt{15})^2}\,\text{ via difference of squares}[/math][math]=\sqrt[3]{16-15}[/math][math]= 1[/math]We observe that [math]\sqrt[3]{4-\sqrt{15}}\cdot\sqrt[3]{4+\sqrt{15}}=1\[/math]
I actually did not see that until the very end of my work.Excellent work above by @Steven G and @red12dog34
Just let me expand on the above (even though it has more detail than Steven's answer) because it took me a while to see this (and I suspect that others may benefit too)...
[math]\sqrt[3]{4-\sqrt{15}}\cdot\sqrt[3]{4+\sqrt{15}}[/math][math]=\sqrt[3]{\left(4-\sqrt{15}\right) \left(4+\sqrt{15}\right)}[/math][math]=\sqrt[3]{4^2-(\sqrt{15})^2}\,\text{ via difference of squares}[/math][math]=\sqrt[3]{16-15}[/math][math]= 1[/math]
Yes, this work is elegant. Work like this is what mathematics an art. Simply beautiful work.We observe that [math]\sqrt[3]{4-\sqrt{15}}\cdot\sqrt[3]{4+\sqrt{15}}=1\Rightarrow\sqrt[3]{4+\sqrt{15}}=\frac{1}{\sqrt[3]{4-\sqrt{15}}}[/math]Let [math]\left(\sqrt[3]{4-\sqrt{15}}\right)^x=t[/math]Then [math] t+\frac{1}{t}=8\Rightarrow t^2-8t+1=0\Rightarrow t_{1,2}=4\pm\sqrt{15}[/math][math]\left(\sqrt[3]{4-\sqrt{15}}\right)^x=4-\sqrt{15}\Rightarrow x=3[/math][math]\left(\sqrt[3]{4-\sqrt{15}}\right)^x=4+\sqrt{15}=\left(4-\sqrt{15}\right)^{-1}\Rightarrow x=-3[/math]