Dear Uncle Colin,

I need to work out values of $a$, $b$ and $c$ that satisfy:

• $abc=14$
• $a+5b+5c=-12$
• $5bc+ab+ac=-3$

How would you do it?

- Some Nasty Equations? Ask… Kolin ((Hey! That’s not my name!))?

Hello, SNEAK, and thanks for your message!

There’s a standard and dull way to do this, and a more exciting way.

### Standard and dull

Rearrange the second equation to make $a$ the subject: $a = -12 - 5b - 5c$.

Substitute this into the two remaining equations:

• $(12 + 5b + 5c)bc = -14$
• $5bc - (b+c)(12 + 5b + 5c) = -3$

Cry a bit, then expand the brackets. Eventually find another variable to eliminate. Solve the resulting cubic. Cry some more. Lament that this is why maths has such a bad reputation. Let’s not do it that way.

### More exciting

I’ve probably oversold this now, haven’t I? Sorry.

I’m going to start by setting $a = 5A$, because I’ve spotted a symmetry.

• $5Abc = 14$, so $Abc = 14/5$
• $5A + 5b + 5b = -12$, so $A + b + c = -12/5$
• $5bc + 5Ac + 5Ab = -3$, so $bc + Ac + Ab = -3/5$.

That might not look, much nicer, but it really is! How come?

Consider the cubic $(A-x)(b-x)(c-x)=0$.

That expands to $Abc - (Ab + bc + cA)x + (A+b+c)x^2 - x^3 = 0$

… and we know about all three of those given coefficients: $\frac{14}{5} + \frac{3}{5}x - \frac{12}{5}x^2 - x^3 = 0$

Or better, $14 + 3x - 12x^2 - 5x^3 = 0$

There’s a fairly obvious solution of $x=1$, so $(1-x)$ is a factor: $(1-x)(14 +17x -5x^2)=0$, or $(1-x)(7-5x)(2-x) = 0$.

The fundamental theorem of arithmetic says that this factorisation must match the original one, so $a/5$, $b$ and $c$ are $1$, $7/5$ and $2$ in any order.

That gives us six sets of solutions:

• $a = 5$, $b = 7/5$ and $c=2$
• $a = 5$, $b = 2$ and $c = 7/5$
• $a = 7$, $b= 1$ and $c=2$
• $a = 7$, $b=2$ and $c=1$
• $a = 10$, $b=7/5$ and $c=2$
• $a = 10$, $b=2$ and $c=7/5$

Are you excited? You ought to be. That’s super-neat.

I hope it helped!

- Uncle Colin