7) Simplify \( \frac{x^3-2x^2-3x+6}{x^3-8} \).

Question

\( \frac{x^3-2x^2-3x+6}{x^3-8} = \frac{x^2(x-2)-3(x-2)}{(x-2)(x^2+2x+4)} = \frac{(x-2)(x^2-3)}{(x-2)(x^2+2x+4)} = \frac{x^2-3}{x^2+2x+4} \), provided \( x \neq 2 \).

Accepted Answer

\( \frac{x^3-2x^2-3x+6}{x^3-8} = \frac{x^2(x-2)-3(x-2)}{(x-2)(x^2+2x+4)} = \frac{(x-2)(x^2-3)}{(x-2)(x^2+2x+4)} = \frac{x^2-3}{x^2+2x+4} \), provided \( x \neq 2 \).

7) Simplify \( \frac{x^3-2x^2-3x+6}{x^3-8} \).

Ответ:

Похожие