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<p><span class="editorial">Alex from Stoke-on-Trent Sixth Form
College, Dapeng Wang from Claremont Fan Court School, Chuyi Yang
from Loughborough High School, Manuele Cavalli-Sforza from the
British School of Manila, Feline Angel from Wootton Upper School
and Chong Ching Tong, Chan Hei Leong, Chen Wei Jian and Ng Yan Shun
from River Valley High School, Singapore all sent in excellent
solutions. The first part of the solution came from the Singapore
group and the second part, with the diagrams, from
Alex.</span>     </p>
<p>If $a + b + c = 4$, $ab + bc + ca = 6$ and $abc = 3$, then</p>
<p>$$\frac {1}{a} +\frac{1}{b} + \frac{1}{c} = \frac {(bc + ac
+ab)}{ abc}= \frac {6}{3} = 2.$$</p>
<p>Also $$\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}=\frac{( c + a
+ b )}{abc}= \frac{4}{3}.$$</p>
<p>In the image below, each side of the coloured squares has been
assigned the lengths, and the area of each rectangle is written
inside the rectangle. The total area of the diagram is the sum of
these areas, and is equivalent to $(a+b+c)^2$ showing that this
expands to: $a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$.</p>
<p>Squaring both sides of the first equation gives $(a+b+c)^2 =
16$. Multiplying each side of the second equation by 2 gives $2ab +
2ac + 2bc = 12$ . Subtracting these last two
equations leaves the sum of the squares so we have $a^2
+ b^2 + c^2 = 4$. </p>
<p><mdo:image height="587" width="551" alt="cube" src="cubediagram.jpg"></mdo:image></p>
 
<table border="0">
<tbody>
<tr>
<td><mdo:image height="232" width="242" src="cube.gif" alt="cube"></mdo:image></td>
<td>
<div>Now think of a cube where each face is cut up in a similar way
splitting the cube into 27 smaller cuboids. The volume of the cube
is $(a + b + c)^3$ . We get the formula for the expansion by adding
27 volumes as shown in the diagrams above and in the table.</div>
<br></br>

<div>The cube has been split into 3 diagrams. The first is the top
layer of the cube, with depth a, so the volumes of each of the
cuboids is the area of the rectangle multiplied by a. The same is
done for the layers below of depth b and c, so the total of the
volumes of the cuboids is equivalent to $(a+b+c)^3$. The expression
labelling each cuboid is the volume of that cuboid.</div>
</td>
</tr>
</tbody>
</table>
 
<table cellspacing="0" cellpadding="5" border="1">
<tbody>
<tr>
<th>Top Layer</th>
<td>$a^2$</td>
<td></td>
<td></td>
<td>$ 2a^2b$</td>
<td>$2a^2c$</td>
<td>$ab^2$</td>
<td>$ac^2$</td>
<td></td>
<td></td>
<td>$2abc$</td>
</tr>
<tr>
<th>Middle Layer</th>
<td></td>
<td>$b^3$</td>
<td></td>
<td>$ a^2b$</td>
<td></td>
<td>$2ab^2$</td>
<td></td>
<td>$2bc^2</td>
<td>$bc^2$</td>
<td>$2abc$</td>
</tr>
<tr>
<th>Bottom Layer</th>
<td></td>
<td></td>
<td>$c^3$</td>
<td></td>
<td>$a^2c$</td>
<td></td>
<td>$2ac^2$</td>
<td>$b^2c$</td>
<td>$2bc^2$</td>
<td>$2abc$</td>
</tr>
</tbody>
</table>
<p>Adding the volumes we get</p>
<p>$(a+b+c)^3 = a^3 + b^3 + c^3+ 3a^2b+3a^2c+3ab^2+3ac^2
+3b^2c+3bc^2+6abc$.</p>
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