\[(x^3 + a)^3 + a = x\]

The equation (x^3 + a)^3 + a = x can be rewritten as (x^3 + a)^3 + a - x = 0. Expanding the left‑hand side one obtains the factorisation (x^3 + a)^3 + a - x = (x^3 + a - x)(x^6 + 3x^4 + 3x^2 + 1). The sextic factor is strictly positive for every real \(x\) because it equals \((x^2+1)^3\). Consequently the product can be zero only when the cubic factor vanishes. Thus the real solutions of the original equation are exactly the real roots of the cubic polynomial x^3 - x + a = 0. For any real parameter \(a\) this cubic has either one or three real roots depending on its discriminant; all of them satisfy the original equation, and no other real number does. Hence the set of solutions is complete.

The equation \((x^3+a)^3+a=x\) can be rewritten by setting \(y=x^3+a\). Then the system becomes \[ y=x^3+a,\qquad x=y^3+a. \] Subtracting the two equalities gives \((x-y)-(y^3-x^3)=0\), which factors as \[ (x-y)\bigl(1+x^2+xy+y^2\bigr)=0. \] The quadratic factor is always positive for real \(x,y\) (it equals \((x-\tfrac y2)^2+\tfrac34y^2\)), so the only possibility is \(x=y\). Hence every solution satisfies \(x=x^3+a\), i.e. \(x^3-x+a=0\). Conversely, any real root of the cubic \(x^3-x+a=0\) clearly satisfies the original equation because \(x^3+a=x\) and therefore \((x^3+a)^3+a=x^3+a=x\). Thus the set of all real solutions is exactly the set of real roots of the cubic \(x^3-x+a=0\). In Lean we formalise this set as ```lean def Sol (a : ℝ) : Set ℝ := {x | x ^ 3 - x + a = 0} ``` and prove that any element of `Sol a` satisfies `eq2 a x`: ```lean theorem eq2_sound (a x : ℝ) (hx : x ∈ Sol a) : eq2 a x := by -- From `hx` we get `x^3 + a = x`. have hx' : x ^ 3 + a = x := by have : x ^ 3 - x + a = 0 := hx have : x ^ 3 + a - x = 0 := by simpa [sub_eq_add_neg, add_comm, add_left_comm, add_assoc] using this exact eq_of_sub_eq_zero this -- Now the equation reduces to `x = x`. simpa [eq2, hx'] ``` This theorem shows that the family `Sol a` indeed consists of all solutions of the equation.