Written by YureQ

Natural number

Recently, while I was tutoring my cousin (2nd grade of high school) I have encountered a following math problem:

Prove that a following expression

\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}

is a natural number.

After some thought (about 15 minutes) I’ve managed to came up with a solution. It is based on the formula for “perfect” square:

(a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}

We can apply it to original expression as follows

\left(\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}\right)^{3}=6\sqrt{3}+10-6\sqrt{3}+10-3\sqrt[3]{\left(6\sqrt{3}+10\right)\left(6\sqrt{3}+10\right)^{2}}+3\sqrt[3]{\left(6\sqrt{3}+10\right)^{2}\left(6\sqrt{3}+10\right)}=20-3\sqrt[3]{8\left(6\sqrt{3}+10\right)}+3\sqrt[3]{8\left(6\sqrt{3}-10\right)}=20-6\left(\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}\ \right)

after substitution

x=\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}

we obtain a 3rd order equation

x^{3}+6x-20=0

It is relatively simple to verify, that only real root of this equation is x=2. So the problem is solved.

I’m not writing it here to boast about my math-solving skills, but about repercussions of this solutions. When my cousin had presented this solution to her teacher she heard that “this is the most complicated way to get the solution”. I was little puzzled. Apparently the “recommended” solution is to observe that

6\sqrt{3}+10=3\sqrt{3}+9+3\sqrt{3}+1=(\sqrt{3})^{3}+3(\sqrt{3})^{2}+3 \sqrt{3} 1^2+1^{3}=(\sqrt{3}+1)^3

and in analogous fashion

6\sqrt{3}-10=(\sqrt{3}-1)^3

in that case we have

\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}=\sqrt[3](\sqrt{3}+1)^3-\sqrt[3](\sqrt{3}-1)^3=\sqrt{3}+1-\sqrt{3}+1=2

It is a great, elegant solution on the condition that you know it. I didn’t and I came up with solution above. What is semi-interesting most of my colleagues from AGH (all academic teachers) came up with the same solution. I started thinking about this recommended solution in the general case. Precisely  how to determine x, y such as

a\sqrt{c}+b=(x \sqrt{c}+y)^3

After quick computation I’ve received a system of equations:

a={x}^{3}c+3\,x{y}^{2}

b=3\,{x}^{2}cy+{y}^{3}

This system has an analytical solution (obtained with Maple)

x=3\,{\frac {as}{8\,s+b}}

y=s

where s is a root of a following polynomial

64\,{s}^{9}-48\,b{s}^{6}+ \left( 27\,{a}^{2}c-15\,{b}^{2} \right) {s}^{3}-{b}^{3}
These roots can be of course computed, but actually that is not the point I am trying to make. If the student (2nd grade of high-school) wants to solve this problem in this way she has to guess the solution or otherwise they are completely screwed. And the teacher is encouraging the guessing approach as the best and simplest one. And we wonder why our students (in University) have difficulties with basic mathematics. Any opinions are welcome.

YureQ

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