I notice that the left-hand side of the equation is a rational expression, and I wonder if I can simplify it by canceling out any common factors. I try to factor the numerator and denominator of the left-hand side. I see that the numerator is a quadratic expression, so I look for two numbers that multiply to -56 and add to -1. I find that -8 and 7 work, so I can write the numerator as $(t-8)(t+7)$. I see that the denominator is already in the form of a linear factor, so I can leave it as $t-8$. I now have  \[\frac{(t-8)(t+7)}{t-8} = \frac{3}{t+5}.\] I see that I can cancel out the common factor of $t-8$ from both sides of the equation, as long as $t \neq 8$. I get  \[t+7 = \frac{3}{t+5}.\] I cross-multiply to get rid of the fractions. I get  \[(t+7)(t+5) = 3.\] I expand the left-hand side using the distributive property. I get  \[t^2 + 12t + 35 = 3.\] I subtract 3 from both sides to get everything on one side of the equation. I get  \[t^2 + 12t + 32 = 0.\] I try to factor the left-hand side again, looking for two numbers that multiply to 32 and add to 12. I find that 8 and 4 work, so I can write the left-hand side as $(t+8)(t+4)$. I set each factor equal to zero and solve for $t$. I get  \[t+8 = 0 \quad \text{or} \quad t+4 = 0.\] I get  \[t = -8 \quad \text{or} \quad t = -4.\] These are the possible values of $t$ that satisfy the equation, but I need to find the greatest one. I see that $-4 > -8$, so the greatest value of $t$ is -4. 
Find the greatest value of $t$ such that  \[\frac{t^2 - t -56}{t-8} = \frac{3}{t+5}.\]