AoPSWiki

Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!

Personal tools

Search

Toolbox

1990 AIME Problems/Problem 2

From AoPSWiki

Problem

Find the value of $(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}$ .

Solution

Suppose that $52+6\sqrt{43}$ is in the form of $(a + b\sqrt{43})^2$ . FOILing yields that $52 + 6\sqrt{43} = a^2 + 43b^2 + 2ab\sqrt{43}$ . This implies that $a$ and $b$ equal one of $\pm1, \pm3$ . The possible sets are $(3,1)$ and $(-3,-1)$ ; the latter can be discarded since the square root must be positive. This means that $52 + 6\sqrt{43} = (\sqrt{43} + 3)^2$ . Repeating this for $52-6\sqrt{43}$ , the only feasible possibility is $(\sqrt{43} - 3)^2$ .

Rewriting, we get $(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3$ . Using the difference of cubes, we get that $[\sqrt{43} + 3\ - \sqrt{43} + 3]\ [(43 + 6\sqrt{43} + 9) + (43 - 9) + (43 - 6\sqrt{43} + 9)]$ $= (6)(3 \cdot 43 + 9) = 828$ .

See also

1990 AIME (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1990_AIME_Problems/Problem_2"

Category: Intermediate Algebra Problems

Add a glimpse of the Art of Problem Solving Forum to your own site!
Click here for details!

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us