Some strange problems struck my mind recently, let me share them with this community. Everyone is welcome to post solutions.

Find the value of this strange infinitely nested radical:

\(\sqrt [ 3 ]{ 6+\sqrt [ 3 ]{ 6+\sqrt [ 3 ]{ 6+\sqrt [ 3 ]{ 6+... } } } }\)

Even stranger :

\(\sqrt [ 3 ]{ 6+\sqrt [ ]{ 6+\sqrt [ 3 ]{ 6+\sqrt [ ]{ 6+\sqrt [ 3 ]{ 6+\sqrt { 6+...} } } } } }\)

But the strangest:

\(\sqrt [ 3 ]{ 6+\sqrt [ ]{ 6-\sqrt [ 3 ]{ 6+\sqrt [ ]{ 6-\sqrt [ 3 ]{ 6+\sqrt { 6-... } } } } } }\)

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## Comments

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TopNewest\(1)\): \(\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\ldots}}}=x \\ \sqrt[3]{6+x}=x \Rightarrow 6+x=x^3 \\ x^3-x-6=0 \\ (x-2)(x^2+2x+3)\Rightarrow x=\boxed{2}\)

\(2)\): \(\sqrt[3]{6+\sqrt{6+\sqrt[3]{6+\sqrt{6+\ldots}}}}=x \\ \sqrt[3]{6+\sqrt{6+x}}=x\Rightarrow \sqrt{6+x}=x^3-6 \\ 6+x=x^6-12x^3+36 \\ x^6-12x^3-x+30=0 \\ \Rightarrow \text{No integral solution} \)

\(3)\) \(\sqrt[3]{6+\sqrt{6-\sqrt[3]{6+\sqrt{6-\ldots}}}}=x \\ \sqrt[3]{6+\sqrt{6-x}}=x\Rightarrow \sqrt{6-x}=x^3-6 \\ 6-x=x^6-12x^3+36 \\ x^6-12x^3+x+30=0 \Rightarrow x=\boxed{2} \)

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for 2, there are two real roots like 1.83 and 2.something

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Yes, but they aren't integral, therefore, I didn't stated them in my solution.

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1) \(2\)

2) Solve \(x^6 - 12x^3 - x + 30 = 0 \)

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