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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, 10 months ago

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

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

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

for 2, there are two real roots like 1.83 and 2.something · 1 year, 10 months ago

Yes, but they aren't integral, therefore, I didn't stated them in my solution. · 1 year, 10 months ago

1) $$2$$

2) Solve $$x^6 - 12x^3 - x + 30 = 0$$ · 1 year, 10 months ago