Cube roots

Algebra Level 3

$\sqrt[3]{ x \sqrt[3]{x \sqrt[3]{ x \sqrt[3]{........ }}}} = ?$

Note - $x \neq 0$

x

x^3

x^{\frac{2}{3}}

x^{ \frac{1}{3} }

\sqrt{ x }

2 solutions

Paul Ryan Longhas
Mar 26, 2015

Let $y = \sqrt[3]{ x \sqrt[3]{x \sqrt[3]{ x \sqrt[3]{........ }}}}$

$=> y^3 = x(\sqrt[3]{ x \sqrt[3]{x \sqrt[3]{ x \sqrt[3]{........ }}}}$

$=> y^3 = xy$

$=> y^3 - xy = 0 => y(y^2 -x)=0$

$=> y = 0$ or $y = \sqrt{x}$

If $x = 0 -----> y = 0$

If $x \in \mathbb{R^{*}} -----> y = \sqrt{x}$

The problem not include that $x \neq 0$ .

@Sakanksha Deo can you add the assumption $x \neq 0$ .

Thank you....... Done

Sakanksha Deo - 6 years, 2 months ago

Hey Sakansha do you go to any coaching class?

Aman Atman - 6 years, 1 month ago

Nice solution! I took a similar approach.

Benry Burfer - 5 years, 8 months ago

Jason Hughes
May 11, 2015

Let $y= \sqrt[3]{ x \sqrt[3]{x \sqrt[3]{ x \sqrt[3]{........ }}}}$ rearrange that as $y= x^{ \frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...}$

The power $x$ is raised to is a infinite geometric series which converges to $\frac{1}{2}$ so y equals $x^{\frac{1}{2}}$ or $\sqrt{x}$ .