Math olympiad basic number theory problem

[timetraveller123]

so this is the question:
let a and b be real numbers such that 0<a<b. Suppose that a³ = 3a -1 and b³ = 3b -1. Find the value of b² -a.
initially my line of thinking was that just solve the equation x³ - 3x +1 = 0
and take the roots which are more than 0 and then after that i got stuck

ok that was the question i also have the full worked solution to it but the thing is that i am almost 100% sure i wouldn't think of such a smart solution in the exam so i want some of you all to solve it post your method to see if anyone has different method.
if you want me to post the solution just ask me to post it below and i would gladly do so. but i am pretty sure many of you should have various other methods to this and no calculator is allowed on this exam and this meant for grade 9 and 10 so please don't use some super complicated methods thanks

 

[PeroK]

Here's how I did it. There are three roots of the equation, call these $c, a, b$, where $c$ is the negative root. From the coefficients:

$c + a + b = 0$ hence $c = -(a+b)$

$ab + ac + bc = -3$ hence (substituting $c$ above) $a^2 + ab + b^2 = 3$

Now, if we let $a = b^2 - \alpha$ we get:

$b(b^3 + (1-2\alpha)b + (3-\alpha)) = 4 - \alpha^2$

Which has a solution of $\alpha = 2$.

 

[timetraveller123]

but if a = 2 then the equation a³ = 3a -1 is not satisfied

 

[PeroK]

but if a = 2 then the equation a³ = 3a -1 is not satisfied

$\alpha = b^2 - a = 2$

 

[timetraveller123]

wait your answer is correct this was the method provided
if t is a solution then t² -2 is also one
then hence a , b , c = a² -2 , b² -2 , c² then by matching the solution based on negative or positive then the answer follows
is this solution super hard to think of and by the way your solution is far more better at least some think i can think of

 

[timetraveller123]

Here's how I did it. There are three roots of the equation, call these $c, a, b$, where $c$ is the negative root. From the coefficients:

$c + a + b = 0$ hence $c = -(a+b)$

$ab + ac + bc = -3$ hence (substituting $c$ above) $a^2 + ab + b^2 = 3$

Now, if we let $a = b^2 - \alpha$ we get:

$b(b^3 + (1-2\alpha)b + (3-\alpha)) = 4 - \alpha^2$

Which has a solution of $\alpha = 2$.

i don't understand how you went from step 3 to 4
please help me thanks

 

[PeroK]

i don't understand how you went from step 3 to 4
please help me thanks

Just a bit of algebra.

 

[timetraveller123]

ya i tried it but not getting what you get pls show your working thanks because i like your method over the given one

 

[PeroK]

ya i tried it but not getting what you get pls show your working thanks because i like your method over the given one

If $a = b^2 - \alpha$, then what is $a^2$?

 

[timetraveller123]

b² - a =x
i don't how to type your a so i used x instead

b² - x =a
substituing in
(b² - x)² + (b² -x)b + b² = 3
x² + b⁴ - 2b²x +b(b² -x) + b² =3
then where do you go from here

 

[PeroK]

b² - a =x
i don't how to type your a so i used x instead

b² - x =a
substituing in
(b² - x)² + (b² -x)b + b² = 3
x² + b⁴ - 2b²x +b(b² -x) + b² =3
then where do you go from here

The next step is:

$b^4 - 2b^2x + b^3-xb + b^2 = 3 - x^2$

Now use $b^3 = 3b - 1$

 

[timetraveller123]

ooh okay i will give it a try
ok i got how to get just realized how dumb i was sorry for that
b(b³ -b(1-2x) +3-x) = 4 - x²
but how do you solve it?

 

[PeroK]

ooh okay i will give it a try
ok i got how to get just realized how dumb i was sorry for that
b(b³ -b(1-2x) +3-x) = 4 - x²
but how do you solve it?

I was trying to get a factor of $b^3 - 3b + 1$ and I saw that $x=2$ does that and solves the equation.

You have a sign wrong there. It should be:

$b(b^3 + b(1-2x) + 3-x) = 4-x^2$

 

[timetraveller123]

oh you about the sign careless
but back to solving i noted that observation but how could you do it rigorously is there a method if there isn't i am okay with this method because it seems fair enough but if there is a method then please do tell me because the original solution has another method of letting b = 2sinΘ which i would never have thought of

 

[PeroK]

oh you about the sign careless
but back to solving i noted that observation but how could you do it rigorously is there a method if there isn't i am okay with this method because it seems fair enough but if there is a method then please do tell me because the original solution has another method of letting b = 2sinΘ which i would never have thought of

The slow way is to solve the quadratic in $x$ and show, by whatever means, that $x=2$ is the only valid solution.

 

[timetraveller123]

ooh okay i will give it a try and get back to you thanks but should i use b as the variable or x as the variable