\(\frac{x^3+8}{x+2}\).
x\(^3\) + 8 = x\(^3\) + 23
recall that
x\(^3\) + y\(^3\) = (x+y)(x\(^2\) - xy + y\(^2\))
x = x, y = 2
x\(^3\) + 8 = (x+2)(x\(^3\) - 2(x) + 22)
\(\frac{x^3+8}{x+2} = \frac{(x+2)(x^3 - 2(x) + 22)}{x+2}\) → x\(^2\) - 2x + 4
x = -2
(-2)\(^2\) - 2(2) + 4 = 4 - 4 + 4
= 4
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