To solve the equation \( \frac{1}{x^2 - x + 1} = \frac{x}{x^2 - x} \) and demonstrate multiplying by the common denominator, we first need to identify the common denominator.
The two denominators in the equation are:
- \( x^2 - x + 1 \)
- \( x^2 - x \)
The common denominator for both is the product of these two expressions. Thus, the common denominator \( D \) is:
\[ D = (x^2 - x + 1)(x^2 - x) \]
Next, we multiply both sides of the equation by this common denominator to eliminate the fractions:
\[ D \cdot \frac{1}{x^2 - x + 1} = D \cdot \frac{x}{x^2 - x} \]
Now, simplifying the left side:
\[ D \cdot \frac{1}{x^2 - x + 1} = (x^2 - x)(1) = x^2 - x \]
For the right side:
\[ D \cdot \frac{x}{x^2 - x} = (x^2 - x + 1)(x) \]
Now our equation simplifies to:
\[ x^2 - x = (x^2 - x + 1)(x) \]
Expanding the right side gives:
\[ x^2 - x = x^3 - x^2 + x \]
Now, we have a polynomial equation:
\[ x^2 - x = x^3 - x^2 + x \]
To bring everything to one side, we rearrange it:
\[ 0 = x^3 - 2x^2 + 2x \]
We can factor this polynomial to make it easier to solve:
\[ 0 = x(x^2 - 2x + 2) \]
The factored form indicates one solution is \( x = 0 \). To find the other solutions, we need to consider the quadratic:
\[ x^2 - 2x + 2 = 0 \]
Calculating the discriminant:
\[ D = b^2 - 4ac = (-2)^2 - 4(1)(2) = 4 - 8 = -4 \]
Since the discriminant is negative, there are no real solutions for this quadratic.
Therefore, the only solution to the original equation is:
\[ \boxed{0} \]
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