1. State the problem. Solve the equation $\sqrt[3]{5x+5}+4=-1$. 2. Formula and important rules. For cube-root equations, use the rule that if $y=\sqrt[3]{u}$ then $y^3=u$. Note that cubing both sides preserves equality for real numbers and does not introduce extraneous solutions because the cube function is bijective on the reals. 3. Isolate the cube root. Subtract 4 from both sides to get $\sqrt[3]{5x+5}=-5$. 4. Apply the cube operation to both sides. Cube both sides using the rule from step 2 to obtain $(\sqrt[3]{5x+5})^3 = (-5)^3$. This simplifies to $5x+5=-125$. 5. Solve the resulting linear equation. Subtract 5 from both sides to get $5x=-130$. Divide both sides by 5: $\frac{5x}{5}=\frac{-130}{5}$. Show cancellation of the common factor 5: $\frac{\cancel{5}x}{\cancel{5}}=\frac{-130}{5}$. Now simplify the right-hand side to get $x=-26$. 6. Check the solution. Substitute $x=-26$ into the original equation: $\sqrt[3]{5(-26)+5}+4=\sqrt[3]{-125}+4$. Compute $\sqrt[3]{-125}=-5$ so we get $-5+4=-1$ which matches the right-hand side. Therefore the solution is correct. 7. Final answer. The solution set is $\{ -26 \}$.