Question

2²-2
자ㅋ
is
1
The simplified from of
n²+32+9​

Answer 1

$\frac{ {x}^{3} - 27}{ {x}^{2} + 3x + 9 } \\ \\ = \frac{ {(x)}^{3} - {(3)}^{3} }{{x}^{2} + 3x + 9} \\ \\ = \frac{(x - 3)( {(x)}^{2} + x(3) + {(3)}^{2}) }{ {x}^{2} + 3x + 9} \\ \\ = \frac{(x - 3)( {x}^{2} + 3x + 9) }{ {x}^{2} + 3x + 9} \\ \\ = x - 3$