Roots Analysis of Cubic Equation $(x-a)^3+(x-b)^3+(x-c)^3=0$

We are asked to determine the nature of the roots for the cubic equation: $$ (x - a)^3 + (x - b)^3 + (x - c)^3 = 0 $$ This equation involves the sum of three cubes. To analyze the roots, we can transform the equation into a simpler form, specifically a depressed cubic equation.

Mathematical Formulation of the Problem

Let's simplify the equation by making a substitution. Consider the substitution $x = y + k$, where $k$ is a constant chosen to eliminate the quadratic term after expansion. A common choice is the average of $a, b, c$: $$ k = \frac{a+b+c}{3} $$ Substituting $x = y + \frac{a+b+c}{3}$ into the equation yields:

$$ \left(y + \frac{a+b+c}{3} - a\right)^3 + \left(y + \frac{a+b+c}{3} - b\right)^3 + \left(y + \frac{a+b+c}{3} - c\right)^3 = 0 $$ Let's define $A' = \frac{b+c-2a}{3}$, $B' = \frac{a+c-2b}{3}$, and $C' = \frac{a+b-2c}{3}$. Note that $A' + B' + C' = 0$. The equation transforms into: $$ (y + A')^3 + (y + B')^3 + (y + C')^3 = 0 $$ Expanding each term: $$ (y^3 + 3y^2A' + 3y(A')^2 + (A')^3) + (y^3 + 3y^2B' + 3y(B')^2 + (B')^3) + (y^3 + 3y^2C' + 3y(C')^2 + (C')^3) = 0 $$ Combine like terms: $$ 3y^3 + 3y^2(A'+B'+C') + 3y((A')^2+(B')^2+(C')^2) + ((A')^3+(B')^3+(C')^3) = 0 $$ Since $A'+B'+C' = 0$, the equation simplifies significantly. Also, recall the identity: if $P+Q+R=0$, then $P^3+Q^3+R^3 = 3PQR$. Applying this with $P=A', Q=B', R=C'$, we get $(A')^3+(B')^3+(C')^3 = 3A'B'C'$. The equation becomes: $$ 3y^3 + 3y((A')^2+(B')^2+(C')^2) + 3A'B'C' = 0 $$ Divide by 3: $$ y^3 + y((A')^2+(B')^2+(C')^2) + A'B'C' = 0 $$ This is a depressed cubic equation of the form $y^3 + py + q = 0$, where: $$ p = (A')^2 + (B')^2 + (C')^2 $$ $$ q = A'B'C' $$ Substituting the expressions for $A', B', C'$ back, we find: $$ p = \frac{1}{3} \left( (a-b)^2 + (b-c)^2 + (c-a)^2 \right) $$ $$ q = \frac{1}{27} (b+c-2a)(a+c-2b)(a+b-2c) $$

Root Nature Analysis

The nature of the roots of the depressed cubic $y^3 + py + q = 0$ depends on the value of $p$. Consider the derivative of $f(y) = y^3 + py + q$, which is $f'(y) = 3y^2 + p$.

• Case 1: $a, b, c$ are not all equal. In this case, at least one pair from $(a,b)$, $(b,c)$, $(c,a)$ is different. This means $(a-b)^2$, $(b-c)^2$, or $(c-a)^2$ is positive. Therefore, $p = \frac{1}{3} \left( (a-b)^2 + (b-c)^2 + (c-a)^2 \right) > 0$. If $p > 0$, then $f'(y) = 3y^2 + p$ is always positive for all real $y$. This indicates that the function $f(y)$ is strictly increasing. A strictly increasing cubic function crosses the x-axis exactly once. Thus, the equation $y^3 + py + q = 0$ has one real root and a pair of complex conjugate imaginary roots. Since the transformation $x = y + k$ simply shifts the roots, the original equation in $x$ also has one real root and two imaginary roots.
• Case 2: $a = b = c$. In this specific case, $p = \frac{1}{3}(0^2+0^2+0^2) = 0$. The original equation becomes $(x-a)^3 + (x-a)^3 + (x-a)^3 = 0$, which simplifies to $3(x-a)^3 = 0$. This equation has $x=a$ as a triple real root.

The question asks about the general nature of the roots. The case where $a, b, c$ are not all equal covers the vast majority of possibilities and represents the general behavior of the equation.

Conclusion on Root Types

Based on the analysis, when $a, b, c$ are not all equal, the cubic equation $(x-a)^3+(x-b)^3+(x-c)^3=0$ yields one real root and two complex conjugate imaginary roots because the associated depressed cubic has $p>0$. The specific case where $a=b=c$ results in a triple real root, but this is a degenerate case.

Therefore, the most accurate general description of the roots is that there is one real root and two imaginary roots.