2003 AMC 10A Problems/Problem 9

Problem

Simplify $\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}$ .

$\mathrm{(A) \ } \sqrt{x}\qquad \mathrm{(B) \ } \sqrt[3]{x^{2}}\qquad \mathrm{(C) \ } \sqrt[27]{x^{2}}\qquad \mathrm{(D) \ } \sqrt[54]{x}\qquad \mathrm{(E) \ } \sqrt[81]{x^{80}}$

Solution 1

$\sqrt[3]{x\sqrt{x}}=\sqrt[3]{(\sqrt{x})^{2}\cdot\sqrt{x}}=\sqrt[3]{(\sqrt{x})^{3}}=\sqrt{x}$ .

Therefore:

$\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}=\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}=\sqrt[3]{x\sqrt{x}}= \sqrt{x} \Rightarrow \boxed{\mathrm{(A)}\ \sqrt{x}}$

Solution 2

We know that $\sqrt[3]{x\sqrt{x}} = \sqrt[3]{x\cdot(x^\frac{1}{2})} = \sqrt[3]{x^\frac{3}{2}} = x^\frac{1}{2}$ We plug this into $\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}$ and get $\sqrt[3]{x\sqrt[3]{x\cdot(x^\frac{1}{2})}}$ . Again, we substitute and get $\sqrt[3]{x\cdot(x^\frac{1}{2})}$ . We substitute one more time and get $x^\frac{1}{2} = \boxed{(A) \sqrt{x}}$ .

Solution 3 (Lame??)

WLOG, plug in some random square number like $9$ or $4$ , and the output you get after simplifying the expression will always be the square root of the number, so the answer is just $\boxed{(A) \sqrt{x}}$ .

~Darth_Cadet

Video Solution

https://www.youtube.com/watch?v=8DMxo2pi1h8 ~David

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search