Algebra Problem of The Day-4

[BigBeachBanana]

[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]

 

[Steven G]

Spoiler

\(\displaystyle 1 + (\sqrt[3] 4+ \sqrt 15)^{2x} =8(\sqrt[3] 4+ \sqrt 15)^x\)
1+u^2 = 8u
\(\displaystyle u = 4 \pm \sqrt 15\)
\(\displaystyle (\sqrt[3] 4+ \sqrt 15)^x =4 \pm \sqrt 15\)

x=3 or \(\displaystyle \dfrac {3\ln(4-\sqrt 15)}{\ln(4+\sqrt 15)}\)

 

[Steven G]

Spoiler

that second solution can be simplified to -3

 

[Steven G]

ln(a-b)=-ln(1/(a-b))(a+b)/a+b)(a+b)) = - ln(a+b)/ln(a^2-b^2)

Spoiler

\(\displaystyle 1 + (\sqrt[3] 4+ \sqrt 15)^{2x} =8(\sqrt[3] 4+ \sqrt 15)^x\)
1+u^2 = 8u
\(\displaystyle u = 4 \pm \sqrt 15\)
\(\displaystyle (\sqrt[3] 4+ \sqrt 15)^x =4 \pm \sqrt 15\)

x=3 or \(\displaystyle \dfrac {3\ln(4-\sqrt 15)}{\ln(4+\sqrt 15)}\)

Oops, my cube root signs need to be a bit longer.

 

[red12dog34]

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]

 

[Deleted member 4993]

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]

Very elegant derivation.....

 

[Cubist]

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)...

We observe that [math]\sqrt[3]{4-\sqrt{15}}\cdot\sqrt[3]{4+\sqrt{15}}=1\[/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]

 

[Steven G]

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]

I actually did not see that until the very end of my work.

 

[Steven G]

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]

Yes, this work is elegant. Work like this is what mathematics an art. Simply beautiful work.