More elegant solution to puzzle question

[Waldo]

There has to be a formula! Has to be! Come on we can do it (well, some one smarter than me can). Would love to see and wrap my head around the solution for this!

[peak4]

There has to be a formula! Has to be! Come on we can do it (well, some one smarter than me can). Would love to see and wrap my head around the solution for this!

 

Here you go then, someone far brighter than me in the 16 century; https://math.vanderbilt.edu/schectex/courses/cubic/

[Diezel]

Here you go then, someone far brighter than me in the 16 century; https://math.vanderbilt.edu/schectex/courses/cubic/

 

Yes, there are several methods to solve cubic functions. But I assume those are not what the puzzle setter intended us to use. You'd have to be a mathematician to know that formula off by heart, or the trigonometric and hyperbolic formulae.

 

---------- Post added 24-12-2016 at 02:42 ----------

 

Yes I agree trial and error seems a little inelegant. I would expect one is meant to just use trial and error, as solving the equation you posted is rather complicated - much beyond my limited capability.

 

I have put it into symbolab however, which gives a step by step of how to solve that equation, which solves it using the rational root theorem and the zero factor principle.

 

https://www.symbolab.com/solver/step-by-step/x%5E%7B3%7D%20-%209x%20-%203240%20%3D%200

 

I'm afraid I get a Security Alert when I try and open your link Robin. I can only assume you are trying to infect me and my PC with eternal mathematic unknowledge

 

I had thought factorisation of some kind might be used. But I have 1(x^3), -9(x) and -3240. Those don't share a whole bunch of common factors so I ruled that out.

Edited December 24, 2016 by Diezel

[Nobby Stiles]

Nothing wrong with trial and error for problems like this.

Slightly easier if you start with 3x, 3(x+1) and 3(x-1)

Then you get 27(x-1)(x)(x+1) = 3240

Or (x-1)(x)(x+1) = 120

Slightly easier to then find three consecutive integers that multiply together to give 120, I.e. 4, 5 and 6 (as an earlier poster said in fact) which gives the answer of 12, 15 and 18. My A level students hate questions like this!

[Diezel]

Nothing wrong with trial and error for problems like this.

Slightly easier if you start with 3x, 3(x+1) and 3(x-1)

Then you get 27(x-1)(x)(x+1) = 3240

Or (x-1)(x)(x+1) = 120

Slightly easier to then find three consecutive integers that multiply together to give 120, I.e. 4, 5 and 6 (as an earlier poster said in fact) which gives the answer of 12, 15 and 18. My A level students hate questions like this!

 

It didn't occur to me to do that but it does indeed make it much easier. Thanks