## Algebra and Trigonometry 10th Edition

Solutions: $x=25$ $x=-29$
$(x+2)^{\frac{2}{3}}=9~~$ (Cube both sides): $[(x+2)^{\frac{2}{3}}]^3=9^3$ $(x+2)^2=729$ $x+2=±27$ $x=-2±27$ $x=-2+27=25$ or $x=-2-27=-29$ Check the solutions: $x=25$: $(25+2)^{\frac{2}{3}}=(\sqrt[3] {27})^2=3^2=9$ $x=25$: $(-29+2)^{\frac{2}{3}}=(\sqrt[3] {-27})^2=(-3)^2=9$