Intermediate Algebra (12th Edition)

$r=-4$
$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $\sqrt[3]{r^2+2r+8}=\sqrt[3]{r^2+3r+12} ,$ raise both sides of the equation to the third power. Then use properties of equality to isolate and solve the variable. Finally, do checking of the solution/s with the original equation. $\bf{\text{Solution Details:}}$ Raising both sides of the equation to the third power results to \begin{array}{l}\require{cancel} \left(\sqrt[3]{r^2+2r+8}\right)^3=\left(\sqrt[3]{r^2+3r+12}\right)^3 \\\\ r^2+2r+8=r^2+3r+12 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} (r^2-r^2)+(2r-3r)=12-8 \\\\ -r=4 \\\\ r=\dfrac{4}{-1} \\\\ r=-4 .\end{array} Upon checking, $r=-4$ satisfies the original equation.