Another fantastic puzzle!
Having been the person who set the puzzle for this month’s “Wrong, but useful“, I had been missing a puzzle this month. Then yesterday I had a great one to solve courtesy of Colin Beveridge (@icecolbeveridge)! See my post here.
Well it turns out that great puzzles can be a bit like busses, as today after reading my post on that puzzle Stuart Price (@sxpmaths) tweeted this little beaut:
Before you read on, have a go. Gonna on, you know you want to!
Check your working!
Ok, well this is what I did:
I had a look, thought about it a little, ran -1, 1, 0 2 and -2 as x to check for an obvious factor and then thought I’d factorise by equating coefficients. I started to expand (ax – b)(cx-d)(ex-f), but then decided to use plusses and have -b/a etc as roots. As I was doing the coefficient of x in my expansion I realised that I could’ve assumed a=b=c=1 as the coefficient of x^3 is 1, but by then I was fairly sure where it was going so followed it through as was to check my feeling was right.
Here is a photo of my working:
I tweeted the pic and Stuart confirmed the solution was correct, and said he’d forgotten to asked for the sum of the sides. In many ways this makes it even nicer, using all the coefficients, however it would probably have also made it easier to spot!
If you have anymore fantastic puzzles, do let me know!
cavmaths 
I found it by expanding (x-a)(x-b)(x-c) into x^3 – (a+b+c) x^2 + (ab+ac+bc)x-abc, which in turn sugests that abc=1 and ab+ac+bc=6. Great puzzle!
Yes, I over complicated it with coefficients of x in brackets!
Product of roots = -d/a = 1 which is the volume of cuboid
Sum of pairs of roots = c/a = 6 which is half of surface area – so surface area = 12
Nice problem