1

Let $a$ and $b$ be such that $0<a<b$. Suppose that $a^3 = 3a-1$ and $b^3 = 3b-1$. Find $b^2 -a$.

nonuser
  • 90,026
vishnu
  • 11
  • 4
    Are you sure that you don't mean $0<a<b$? – ajotatxe Apr 16 '17 at 10:05
  • oh ya that's what i meant thanks and how do i get a name instead of user#### like you thanks for that too – vishnu Apr 16 '17 at 10:08
  • @user437260 just change your name. – DHMO Apr 16 '17 at 10:13
  • 2
    See https://math.stackexchange.com/questions/2157643/how-can-i-solve-the-equation-x3-x-1-0/2157645#2157645 – lab bhattacharjee Apr 16 '17 at 10:15
  • Adapting the tricks in the thread lab bhattacharjee linked to (his own answer in particular, +1) may be most accessible route to you. If you run into difficulties, you can take a peek at this thread, where we give the zeros of $x^3-3x+1$. Anyway, using the triplication formula for cosines leads to a clean closed form :-) – Jyrki Lahtonen Apr 16 '17 at 10:23
  • @user437260 Go to https://math.stackexchange.com/users/edit/current , then change display name. Note: you can only change your display name once in 30 days. – wythagoras Apr 16 '17 at 10:57
  • @lab bhattacharjee i know the solution using trig substitution it is let x = 2siny but what other methods are there – vishnu Apr 16 '17 at 11:40

2 Answers2

1

Since $0<a<b$ and $a^3=3a-1$ and $b^3=3b-1$, then we need to find the two positive real roots of $x^3-3x+1=0$ with $a$ being the smaller root and $b$ being the bigger root.

Following Cardano's solution, if we let $x=u+v$ then $$ \begin{align*} x^3-3x+1&=0\\ (u+v)^3-3(u+v)+1&=0\\ u^3+3u^2v+3uv^2+v^3-3u-3v+1&=0\\ u^3+v^3+1+3u^2v-3u+3uv^2-3v&=0\\ (u^3+v^3+1)+3(u+v)(uv-1)&=0\\ \end{align*} $$ So we want to find the simultaneous solution to the following system: $$ \begin{align*} u^3+v^3&=-1\\ uv&=1. \end{align*} $$ By substituting $u=\frac1v$ into the first equation we have $$ \frac1{v^3}+v^3=-1\iff v^6+v^3+1=0. $$ Then the quadratic formula tells us $$v^3=\frac{-1+i\sqrt{3}}{2}=e^{2\pi i/3}\iff v=e^{2\pi i/9}\implies u=e^{-2\pi i/9}$$ or $$v^3=\frac{-1-i\sqrt{3}}{2}=e^{4\pi i/3}\iff v=e^{4\pi i/9}\implies u=e^{-4\pi i/9}.$$ So the two real positive solutions are $$ \begin{align*} a&=e^{-4\pi i/9}+e^{4\pi i/9}=2\cos\left(\frac{4\pi}{9}\right)\\ b&=e^{-2\pi i/9}+e^{2\pi i/9}=2\cos\left(\frac{2\pi}{9}\right)\\ \end{align*} $$ and we see that $$ \begin{align*} b^2-a&=4\cos^2\left(\frac{2\pi}{9}\right)-2\cos\left(\frac{4\pi}{9}\right)\\ &=4\cos^2\left(\frac{\frac{4\pi}{9}}{2}\right)-2\cos\left(\frac{4\pi}{9}\right)\\ &=4\left(\frac{1+\cos\left(\frac{4\pi}{9}\right)}{2}\right)-2\cos\left(\frac{4\pi}{9}\right)\\ &=2+2\cos\left(\frac{4\pi}{9}\right)-2\cos\left(\frac{4\pi}{9}\right)\\ &=2. \end{align*} $$

Laars Helenius
  • 7,965
  • 1
  • 23
  • 35
  • pretty sure the junior olympiad doesn't need such an advanced solution but thanks it was insightful – vishnu Apr 19 '17 at 08:37
  • You need to be able to solve a cubic, know that $e^{i\theta}=\cos\theta+i\sin\theta$, and a basic half-angle trig identity. Putting it all together would be tough, but the pieces and parts don't go beyond a precalculus level class. – Laars Helenius Apr 19 '17 at 11:43
  • i fully understand the solution and how to solve using cardano's method but what i am saying is that this Olympiad specifically states it doesn't require calculus and the use of complex numbers but such methods will be accepted if used hence this is a brilliant solution but there are simpler solutions like trig substitution and i really loved your solution thanks again – vishnu Apr 19 '17 at 13:32
1

Substituting $a=2a'$ and $b=2b'$ and dividing the equations by $2$ we have $$4a'^3-3a'=4b'^3-3b'=-1/2.$$ Since $4\cos^3x-3\cos x=\cos 3x$ for all $x$, we see that $\{\cos 40^o,\cos 160^o,\cos 280^o\}$ is the set of solutions to $4y^3-3y=\cos 120^o=-1/2 .$

So $a=2\cos 280^o=2\cos 80^o$ and $b=2\cos 40^o.$ The rest is in the last part of the A by Laars Helenius.

DanielWainfleet
  • 57,985