Solve the equation x³ − 6x² + 11x − 6 = 1.

Step-by-step solution

1. 1

   Move everything to one side and test for rational roots. The equation is $x^3 - 6x^2 + 11x - 7 = 0$, so the rational candidates are the divisors of $7$:

   $f(1) = -1, \quad f(-1) = -25, \quad f(7) = 119, \quad f(-7) = -721$

   None vanishes. Note how the change from $-6$ to $-7$ destroys the tidy factorisation the original left side had — $x = 1$ now misses by exactly $1$.

2. 2

   Exploit the structure of the left-hand side. The polynomial that appears is the classic product of three consecutive linear factors:

   $x^3 - 6x^2 + 11x - 6 = (x-1)(x-2)(x-3)$

   So the equation is really $(x-1)(x-2)(x-3) = 1$ — three consecutive integers-apart factors whose product is $1$.

3. 3

   Centre the substitution on the middle factor. Let $y = x - 2$, so the three factors become $y+1$, $y$, $y-1$:

   $(y+1)\,y\,(y-1) = y\left(y^2-1\right) = y^3 - y$

   $\Longrightarrow \quad y^3 - y = 1, \quad \text{i.e.} \quad y^3 - y - 1 = 0$

   Shifting to the symmetry point of the factors is what removes the quadratic term for free — no Cardano depression algebra needed.

4. 4

   Count the real roots of the depressed cubic. For $y^3 + py + q$ with $p = -1$, $q = -1$, the discriminant is

   $\Delta = -4p^3 - 27q^2 = 4 - 27 = -23 < 0$

   so $y^3 - y - 1 = 0$ has exactly one real root and a complex conjugate pair. The real root is the plastic number $\rho$.

5. 5

   Evaluate the real root. Cardano's formula gives

   $\rho = \sqrt[3]{\frac12 + \sqrt{\frac{23}{108}}} + \sqrt[3]{\frac12 - \sqrt{\frac{23}{108}}} = 1.3247179572\ldots$

   Newton's method on $y^3 - y - 1$ from $y_0 = 1.3$ returns the same value to 39 digits, so the two routes agree.

6. 6

   Undo the substitution and check. Since $x = y + 2$,

   $x = 2 + \rho \approx 3.3247179572$

   Substituting into the original: $(x-1)(x-2)(x-3) \approx (2.3247)(1.3247)(0.3247) = 1.0000$, and evaluating $x^3-6x^2+11x-6$ at 40-digit precision gives $0.99999999999999999999999999999999999999$. Only one real solution exists.