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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
1 year 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 · 1 year ago

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

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

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1) \(2\)

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

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