#1. In this problem, we will find the real roots x of the equation x^3=3x+1 . a)

Question

#1. In this problem, we will find the real roots x of the equation x^3=3x+1.
a) Substitute s+t for x, where s and t are new variables such that st=1. Show that the equation simplifies to s^3+t^3=1.
b) Use the relation t=1/s to obtain an equation with s as the only variable. Multiply through by s^3 to obtain an equation that is quadratic in s^3 and solve. Notice that whichever of the two solutions for s^3 you choose in the quadratic formula, t^3 becomes the other solution, so it doesn't matter which you choose.
c) Now that you have s^3 and t^3, find values for s and t, and thus for x, which was s+t. Remember that there are three cube roots each for s^3 and t^3, and for each cube root s of s^3, pair it with the cube root t of t^3 such that st=1 holds, as we required before. This will give three values for x. Simplify each value of x so that it is clear it is a real number.

Need help with (b)(c)

Explanation / Answer

1) x^3 = 3x + 1

x = s + t

st = 1

(s + t)^3 = 3(s + t) + 1

=> s^3 + t^3 + 3s^2*t + 3st^2 = 3(s + t) + 1

=> s^3 + t^3 + 3st(s + t) = 3(s + t) + 1

=> s^3 + t^3 + 3(1)(s + t) = 3(s + t) + 1

=> s^3 + t^3 = 1

2) t = 1/s

=> s^3 + (1/s)^3 = 1

=> s^3 + 1/s^3 = 1

=> s^6 - s^3 + 1 = 0

=> (s^3)^2 - (s^3) + 1 = 0

s^3 = ( 1 +- sqrt(1 - 4) ) / 2

=> s^3 = ( 1+- sqrt(-3) ) / 2

If s^3 = (1 + sqrt(-3) ) / 2,

then t^3 = 1 / s^3 = 2 / (1 + sqrt(-3) ) = 2*(1 - sqrt(-3) ) / (1 + sqrt(-3) )*(1 - sqrt(-3) )

= 2(1 - sqrt(-3)) / (1-(-3))

= ( 1 - sqrt(-3) ) / 2

and if s^3 = (1 - sqrt(-3) ) / 2, then t^3 = (1 + sqrt(-3) ) / 2

Let s^3 = (1 + sqrt(-3) ) / 2 and t^3 = ( 1 - sqrt(-3) ) / 2

3) s^3 = (1 + sqrt(-3) ) / 2 = (1)e^(i*pi/3)

=> s^3 = e^(i*pi/3) = e^(i*7pi/3) = e^(i*13pi/3)

=> s = e^(i*pi/9) = e^(i*7pi/9) = e^(i*13pi/9)

t^3 = (1 - sqrt(-3) ) / 2 = (1)e^(i*5pi/3)

=> t^3 = e^(i*5pi/3) = e^(i*11pi/3) = e^(i*17pi/3)

=> t = e^(i*5pi/9) = e^(i*11pi/9) = e^(i*17pi/9)

Pair s and t such that st = 1

(e^(ix) * e^(iy) = e^(ix + iy) )

s = e^(i*pi/9), t = e^(i*17pi/9)

s = e^(i*7pi/9), t = e^(i*11pi/9)

s = e^(i*13pi/9), t = e^(i*5pi/9)

x = s + t = e^(i*pi/9) + e^(i*17pi/9) = cos(pi/9) + cos(17pi/9) = 1.879385

x = e^(i*7pi/9) + e^(i*11pi/9) = -1.532

x = e^(i*13pi/9) + e^(i*5pi/9) = -0.347296

x = 1.879385 (or) x = -1.532 (or) x = -0.347296

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