## College Algebra 7th Edition

$|x^{3}+\frac{1}{4x^{3}}|$
We distribute the squared term, simplify, factor, and simplify again. We take care to remember that $\sqrt{x^2}=|x|$ (not just $x$): $\sqrt{1+(x^{3}-\frac{1}{4x^{3}})^{2}}=\sqrt{1+x^{6}-\frac{2x^{3}}{4x^{3}}+\frac{1}{16x^{6}}}=\sqrt{1+x^{6}-\frac{1}{2}+\frac{1}{16x^{6}}}=\sqrt{x^{6}+\frac{1}{2}+\frac{1}{16x^{6}}} =\sqrt{(x^{3}+\frac{1}{4x^{3}})^{2}}=|x^{3}+\frac{1}{4x^{3}}|$