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Answer:

[tex]\frac{x+2}{3x-1}[/tex]

Explanation:

→ [tex]\frac{(x^2-9)(x+2)}{(3x^2 +8x -3)(x-3)}[/tex]

→ [tex]\frac{\left(x+3\right)\left(x-3\right)\left(x+2\right)}{\left(3x^2+8x-3\right)\left(x-3\right)}[/tex]

→ [tex]\frac{\left(x+3\right)\left(x+2\right)}{3x^2+8x-3}[/tex]

→ [tex]\frac{\left(x+3\right)\left(x+2\right)}{\left(3x-1\right)\left(x+3\right)}[/tex]

→ [tex]\frac{x+2}{3x-1}[/tex]

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