Answer
$g(3)=-\dfrac{17}{48}$
$g(-2)=\dfrac{2}{7}$
$g(1)=-\dfrac{3}{8}$
Work Step by Step
To find $g(3), g(-2), \text{ and } g(1)$, susbtitute 3, -2, and 1, respectively, to $x$ to have:
$g(3)=\dfrac{3^2+8}{3^3-25(3)}=\dfrac{9+8}{27-75}=\dfrac{17}{-48}=-\dfrac{17}{48}
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\\g(-2)=\dfrac{(-2)^2+8}{(-2)^3-25(-2)}=\dfrac{4+8}{-8+50}=\dfrac{12}{42}=\dfrac{2}{7}
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\\g(1)=\dfrac{1^2+8}{1^3-25(1)}=\dfrac{1+8}{1-25}=\dfrac{9}{-24}=-\dfrac{3}{8}$
