## Algebra and Trigonometry 10th Edition

$x=±\sqrt {14}$
$(x^2-5)^{\frac{3}{2}}=27~~$ (Square both sides) $[(x^2-5)^{\frac{3}{2}}]^2=27^2$ $(x^2-5)^3=729$ $x^2-5=\sqrt[3] {729}$ $x^2-5=9$ $x^2=14$ $x=±\sqrt {14}$ Check the solutions: $x=\sqrt {14}$ $[(\sqrt {14})^2-5]^{\frac{3}{2}}=(14-5)^{\frac{3}{2}}=9^{\frac{3}{2}}=(\sqrt 9)^3=3^3=27$ $x=-\sqrt {14}$ $[(-\sqrt {14})^2-5]^{\frac{3}{2}}=(14-5)^{\frac{3}{2}}=9^{\frac{3}{2}}=(\sqrt 9)^3=3^3=27$