Answer
$x^{45}$
Work Step by Step
RECALL:
(i) $\dfrac{a^m}{a^n} = a^{m-n}, a\ne0$
(ii) $(a^m)^n = a^{mn}$
Use rule (i) above to obtain:
$=(x^{4-(-11)})^3
\\=(x^{4+11})^3
\\=(x^{15})^3$
Use rule (ii) above to obtain:
$=x^{15\cdot3}
\\=x^{45}$
