## Precalculus: Mathematics for Calculus, 7th Edition

a) ($x^{2}$)$^{3}$ = $x^{2\times3}$ = $x^{6}$ $\ne$ $x^{5}$ b) ($2x^{4}$)$^{3}$ = $2^{3}$ $\times$ $x^{4\times3}$ = 8$x^{12}$ c) $\sqrt (4a^{2}$) = $\sqrt 4$ $\times$ $\sqrt a^{2}$ = 2a d) One way to see that these two numbers are different is by comparing their squares: ($\sqrt (a^{2}+4$))$^{2}$ = $a^{2}+4$ $\ne$ (a+2)$^{2}$ = a$^{2}$ + 4 + 4a