Answer
$\dfrac{1}{d^3}$
Work Step by Step
RECALL:
(i) $a^m \cdot a^n = a^{m+n}$
(ii) $a^{-m} = \dfrac{1}{a^m}, a\ne0$
Use the rules above to have:
$=d^{14-17}
\\=d^{-3}
\\=\dfrac{1}{d^3}$
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