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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
2 years 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 - 2 years ago

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

Dev Sharma - 2 years ago

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

Akshat Sharda - 2 years ago

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1) \(2\)

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

Dev Sharma - 2 years ago

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