## College Algebra (6th Edition)

Solution set as $x$-intercepts = $\{ 1,729\}$ The graph of equation $y=x^{\frac{1}{3}}+2x^{\frac{1}{6}}-3$ is shown in Graph$(e)$
$y=x^{\frac{1}{3}}+2x^{\frac{1}{6}}-3$ To find $x$ -intercept, $y=0$ $x^{\frac{1}{3}}+2x^{\frac{1}{6}}-3=0$ This equation is equivalent to $(x^{\frac{1}{6}})^{2}+2x^{\frac{1}{6}}-3=0$ Let $(x^{\frac{1}{6}}) = u$ $u^{2}+2u-3=0$ By factoring, $(u-1)(u+3)=0$ $u=1$ or $u=-3$ Let $u=1$ Replace $u= (x^{\frac{1}{6}})$ $(x^{\frac{1}{6}}) =1$ Raise to the power $6$ on both sides $(x^{\frac{1}{6}}) ^6=(1)^{6}$ $x=1$ Let $u=-3$ $(x^{\frac{1}{6}}) =-3$ $(x^{\frac{1}{6}}) ^6=(-3)^{6}$ $x=729$ Solution set as $x$-intercepts = $\{ 1,729\}$ The graph of equation $y=x^{\frac{1}{3}}+2x^{\frac{1}{6}}-3$ is shown in Graph$(e)$