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 :

6+6+6+6+6+6+...333\sqrt [ 3 ]{ 6+\sqrt [ ]{ 6+\sqrt [ 3 ]{ 6+\sqrt [ ]{ 6+\sqrt [ 3 ]{ 6+\sqrt { 6+...} } } } } }

But the strangest:

6+6−6+6−6+6−...333\sqrt [ 3 ]{ 6+\sqrt [ ]{ 6-\sqrt [ 3 ]{ 6+\sqrt [ ]{ 6-\sqrt [ 3 ]{ 6+\sqrt { 6-... } } } } } }

#Algebra #Strangequestions

Note by Swapnil Das
5 years, 6 months ago

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1 vote

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Comments

1)1): 6+6+6+…333=x6+x3=x⇒6+x=x3x3−x−6=0(x−2)(x2+2x+3)⇒x=2\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)2): 6+6+6+6+…33=x6+6+x3=x⇒6+x=x3−66+x=x6−12x3+36x6−12x3−x+30=0⇒No integral solution\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)3) 6+6−6+6−…33=x6+6−x3=x⇒6−x=x3−66−x=x6−12x3+36x6−12x3+x+30=0⇒x=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 x=\boxed{2}

Akshat Sharda - 5 years, 6 months ago

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

Dev Sharma - 5 years, 6 months ago

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

Akshat Sharda - 5 years, 6 months ago

1) 22

2) Solve x6−12x3−x+30=0x^6 - 12x^3 - x + 30 = 0

Dev Sharma - 5 years, 6 months ago
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