Solve the equation x^3 - 3x^2 + 3x - 1 = 0.

Step-by-step solution

1. 1

   Read the coefficients as a binomial row. The absolute values $1,\ 3,\ 3,\ 1$ are row 3 of Pascal's triangle, and the signs alternate. That is exactly the pattern of

   $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$

2. 2

   Match a and b. Taking $a = x$ and $b = 1$ reproduces every term: $x^3 - 3x^2(1) + 3x(1)^2 - 1^3$. Hence

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

   Spotting this saves the whole rational-root-and-divide routine.

3. 3

   Solve the collapsed equation. A cube is zero only when its base is zero:

   $(x-1)^3 = 0 \quad\Longrightarrow\quad x - 1 = 0 \quad\Longrightarrow\quad x = 1$

4. 4

   Describe the multiplicity. The root $x = 1$ has multiplicity $3$, so there is only one distinct solution even though a cubic has three roots counted with multiplicity. Both $f(1) = 0$ and $f'(1) = 3(1)^2 - 6(1) + 3 = 0$, and $f''(1) = 6 - 6 = 0$ — the first three derivatives vanishing is the algebraic signature of a triple root.

5. 5

   Verify and note the graph. $1 - 3 + 3 - 1 = 0$. Because the multiplicity is odd, the curve crosses the $x$-axis at $x = 1$ rather than touching it, but it does so with zero slope, producing the characteristic flat inflection there.