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Strange problems!

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:

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

Note by Swapnil Das
10 months, 1 week ago

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

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

\(3)\) \(\sqrt[3]{6+\sqrt{6-\sqrt[3]{6+\sqrt{6-\ldots}}}}=x \\ \sqrt[3]{6+\sqrt{6-x}}=x\Rightarrow \sqrt{6-x}=x^3-6 \\ 6-x=x^6-12x^3+36 \\ x^6-12x^3+x+30=0 \Rightarrow x=\boxed{2} \) Akshat Sharda · 10 months, 1 week ago

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@Akshat Sharda for 2, there are two real roots like 1.83 and 2.something Dev Sharma · 10 months, 1 week ago

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@Dev Sharma Yes, but they aren't integral, therefore, I didn't stated them in my solution. Akshat Sharda · 10 months, 1 week ago

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1) \(2\)

2) Solve \(x^6 - 12x^3 - x + 30 = 0 \) Dev Sharma · 10 months, 1 week ago

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