By
Yussof
MF6B (2013)
One
Thursday morning, Mr Lord came dashing through my form class and presented us
with a maths problem that was given to him by Dr O’Neill. I looked at it and
thought it was challenging so I gave it a go.
Given:
\(a + b + c = 0\), prove that: \(\left( {\frac{{b - c}}{a} + \frac{{c - a}}{b}
+ \frac{{a - b}}{c}} \right)\left( {\frac{a}{{b - c}} + \frac{b}{{c - a}} +
\frac{c}{{a - b}}} \right) = 9\)
When
I first looked at this problem, I thought it would be wise to expand the two
brackets straight on and tried to simplify it but I ended up having something
really messy so I tried a different approach.
Let:
\(x = \frac{{b - c}}{a},\) \(y = \frac{{c - a}}{b},\) \(z = \frac{{a - b}}{c}\)
Therefore
we get:
\[ =
\left( {x + y + z} \right)\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}
\right)\]
Expanding
these two brackets:
\[ = 3 +
\frac{{z + y}}{x} + \frac{{z + x}}{y} + \frac{{x + y}}{z} \cdot \cdot
\cdot \cdot \cdot
\cdot \cdot \cdot \left(
* \right)\]
Now let
us simplify: \(\begin{array}{c} = \frac{{z + y}}{x}\\ = \frac{1}{x}\left( {z +
y} \right)\\ = \left( {\frac{a}{{b - c}}} \right)\left( {\frac{{a - b}}{c} +
\frac{{c - a}}{b}} \right)\\ = \left( {\frac{a}{{b - c}}} \right)\left[ {\frac{{b\left(
{a - b} \right) + c\left( {c - a} \right)}}{{bc}}} \right]\\ = \left(
{\frac{a}{{b - c}}} \right)\left( {\frac{{ab - {b^2} + {c^2} - ac}}{{bc}}}
\right)\\ = \left( {\frac{a}{{b - c}}} \right)\left( {\frac{{ab - ac - \left(
{{b^2} - {c^2}} \right)}}{{bc}}} \right)\\ = \left( {\frac{a}{{b - c}}}
\right)\left( {\frac{{a\left( {b - c} \right) - \left( {b - c} \right)\left( {b
+ c} \right)}}{{bc}}} \right)\\ = \left( {\frac{a}{{b - c}}} \right)\left( {b -
c} \right)\left( {\frac{{a - b - c}}{{bc}}} \right)\\ = \left( a \right)\left(
{\frac{{a - b - c}}{{bc}}} \right)\end{array}\)
Using \(a
+ b + c = 0\) so, \(a = - b
- c\) : \(\begin{array}{c} = \left( a \right)\left( {\frac{{a + a}}{{bc}}}
\right)\\ = \left( a \right)\left( {\frac{{2a}}{{bc}}} \right)\\ = \left(
{\frac{{2{a^2}}}{{bc}}} \right)\end{array}\)
By
symmetry, we can note that:
\[\frac{{z
+ x}}{y} = \frac{{2{b^2}}}{{ac}}\] & \(\frac{{x + y}}{z} =
\frac{{2{c^2}}}{{ab}}\)
Therefore
going back to equation (*) from above: \(\begin{array}{c} = 3 +
\frac{{2{a^2}}}{{bc}} + \frac{{2{b^2}}}{{ac}} + \frac{{2{c^2}}}{{ab}}\\ = 3 +
\left( {\frac{{2{a^3} + 2{b^3} + 2{c^3}}}{{abc}}} \right)\\ = 3 + 2\left(
{\frac{{{a^3} + {b^3} + {c^3}}}{{abc}}} \right)\end{array}\)
Note from
here onwards I did get stuck for a while because I could not simplify \({a^3} +
{b^3} + {c^3}\) until I found the formula for sum of three cubes# which is:
\[{a^3} +
{b^3} + {c^3} = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab -
bc - ac} \right) + 3abc\]
And since
\(a + b + c = 0\) , \({a^3} +
{b^3} + {c^3} = 3abc\).
Going
back to the equation: \(\begin{array}{c} = 3 + 2\left( {\frac{{{a^3} + {b^3} +
{c^3}}}{{abc}}} \right)\\ = 3 + 2\left( {\frac{{3abc}}{{abc}}} \right)\\ = 3 +
2\left( 3 \right)\\ = 9\left( {{\rm{QED}}} \right)\end{array}\)
