In addition to Ï/4 (pretty clear, through substitution), you can see that there must be two other real solutions by finding the extrema. Taking the derivative, 3x^2 + 2x - 1 = 0 when x = -1 or x = 1/3.
The other two answers are about -1.52 and -0.264; I don't see any way that these are related to Pi.
Ed: Of course! kb's solution is correct - just factor out (x-Ï/4) and you can plug the rest into the quadratic equation.
Answers & Comments
Verified answer
Since x = π/4 is a root, let's substitute x = u + π/4.
(u + π/4)³ + (u + π/4)² - (u + π/4) = π³/64 + π²/16 - π/4.
==> u³ + (1 + 3π/4) u² + (3π²/16 + π/2 - 1)u = 0
==> u (u² + (1 + 3π/4) u + (3π²/16 + π/2 - 1)) = 0
So, u = 0 or
u = [-(1 + 3π/4) [+ or -] sqrt((1 + 3π/4)² - 4(3π²/16 + π/2 - 1))]/2
= [-(1 + 3π/4) [+ or -] sqrt(5 - π/2 - 3π²/16)]/2.
Since x = u + π/4,
u = π/4 or u = π/4 + [-(1 + 3π/4) [+ or -] sqrt(5 - π/2 - 3π²/16)]/2.
I hope that helps!
In addition to Ï/4 (pretty clear, through substitution), you can see that there must be two other real solutions by finding the extrema. Taking the derivative, 3x^2 + 2x - 1 = 0 when x = -1 or x = 1/3.
The other two answers are about -1.52 and -0.264; I don't see any way that these are related to Pi.
Ed: Of course! kb's solution is correct - just factor out (x-Ï/4) and you can plug the rest into the quadratic equation.
x³ + x² - x = ϳ/64 + ϲ/16 - Ï/4
<=> (x³ -ϳ/64 )+(x²-ϲ/16 )-(x- Ï/4)=0
<=> (x- Ï/4)(x²+x.Ï/4 +ϲ/16)+ (x- Ï/4)(x+ Ï/4)-(x- Ï/4)=0
<=> (x- Ï/4)(x²+x.Ï/4 +ϲ/16 +x+ Ï/4-1) = 0
<=> x- Ï/4 = 0 or x²+x.Ï/4 +ϲ/16 +x+ Ï/4-1=0
=> x= Ï/4
Compare
x³ + x² - x = (Ï/4)³ + (Ï/4)² - (Ï/4)
x = Ï/4
One real root is Ï/4,
The other roots can be found by synthetic division.