Question

divide.
\\(\frac{x - x^2}{x^3 - 1} \div \frac{x}{x^2 + x + 1}\\)
\\(\frac{x - x^2}{x^3 - 1} \div \frac{x}{x^2 + x + 1} = \square\\) (simplify your answer.)

Explanation:

Step1: Rewrite division as multiplication

To divide by a fraction, multiply by its reciprocal. So we have:
$$\frac{x - x^2}{x^3 - 1} \div \frac{x}{x^2 + x + 1} = \frac{x - x^2}{x^3 - 1} \times \frac{x^2 + x + 1}{x}$$

Step2: Factor numerators and denominators

• Factor \(x - x^2\): \(x - x^2 = -x(x - 1)\)
• Factor \(x^3 - 1\) (using difference of cubes: \(a^3 - b^3=(a - b)(a^2+ab + b^2)\)): \(x^3 - 1=(x - 1)(x^2 + x + 1)\)

Substitute these factorizations into the expression:
$$\frac{-x(x - 1)}{(x - 1)(x^2 + x + 1)} \times \frac{x^2 + x + 1}{x}$$

Step3: Cancel common factors

• Cancel \((x - 1)\) from numerator and denominator.
• Cancel \(x\) from numerator and denominator.
• Cancel \((x^2 + x + 1)\) from numerator and denominator.

After canceling, we are left with \(-1\).

Answer:

\(-1\)