Associated Topics || Dr. Math Home || Search Dr. Math

### Competition Questions

```
Date: 25 Feb 1995 16:19:42 -0500
From: Anonymous
Subject: math problems

Dear Dr. Math,

Do you also answer questions from teachers?  We are trying to
prepare students for competitions and have (at present) two problems
that we cannot ourselves work!  Here they are:

[1+{28-6(3)^(1/2)}^1/2]^2

The coefficient of the third term of
(8x-y)^(1/3) when simplified is _____?

Without math symbols, the first problem is hard to write; the 3 is
under a radical as is the entire expression in braces.

Hope you can help.  We would appreciate an answer by March 1, if
possible.

Thank you,    Janet Ramser, Clarksville,TN

IN%"ramserj@ten-nash.ten.k12.tn.us"
```

```
Date: 26 Feb 1995 01:53:15 -0500
From: Dr. Ken
Subject: Re: math problems

Hello there!

These are fantastic problems!  Actually, I guess I can only say that the
first one is a fantastic problem, since I haven't really looked at the
second one yet.  But I'll see if I can help with the first one:

The key to this one will be simplifying the term Sqrt{28 - 6 Sqrt{3}}.
Here's how you go about eliminating the nested radicals.  Let
x    = Sqrt{28 - 6*Sqrt{3}}.  Then,
x^2 = 28 - 6*Sqrt{3}.

Now you can proceed in one of two directions.  You can either say, "Hey,
28 is 1 + 27" and right away say that
28 - 6*Sqrt{3} = 27 - 6*Sqrt{3} + 1
= (3*Sqrt{3} - 1)^2, so
x = 3*Sqrt{3} - 1.

Then you have

(1 + 3*Sqrt{3} - 1)^2, which is more easily seen to be 27.

On the other hand (and this is the way I solved this problem), you can
say "okay, x^2 = 28 - 6*Sqrt{3}, so let's try to find something that you
can square to get 28 - 6*Sqrt{3}."

So do this:  take two new variables, a and b, and say
x^2 = (a + b*Sqrt{3})^2
= a^2 + 3b^2 + 2ab*Sqrt{3}.

Then we get a^2 + 3b^2 = 28, and 2ab = 6.

This is a system of equations that you can solve to find a and b, and
once you've found a and b, you can find x to be 3*Sqrt{3} - 1.

As for your second problem.  I asked Professor Charles Grinstead in the
math department here at Swarthmore, and he said this:

The only thing that comes to mind is the Binomial Theorem.  If
one applies this theorem to the given example, one obtains:

(8x-y)^(1/3) = 2x^(1/3)(1-(y/8x))^(1/3)
= 2x^(1/3)(1 + (1/3)(-y/8x) + ((1/3)(-2/3)/2!)(-y/8x)^2 + ...)

That makes a lot of sense to me now.  Do you folks know the binomial
formula?  It says that for any real number, p, and for x such that -1 < x < 1,

(1 + x)^p = 1 + [p:1]*x + [p:2]*x^2 + [p:3]*x^3 + ...,

Where [p:k] is the binomial coefficient defined by

(p)(p-1)(p-2)...(p-k+2)(p-k+1)
[p:k] =  -------------------------------------
k!

So that's what comes to mind when attacking this problem.  Given the wording
of the problem though, I'm not surprised that you were confused by it!

-Ken "Dr." Math

```
Associated Topics:
High School Basic Algebra

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search