Question

Simplify: $\bigg (\frac{b^{-3} \cdot b^7 \cdot (b^{-1})^2} {(-b)^{2} \cdot (b^2)^{3}} \bigg )^{-2}$

Answer 1

$= ( \frac{b { }^{ - 3} \times b {}^{7} \times b {}^{ - 2} }{b {}^{2} \times b {}^{6} } ) {}^{ - 2}$
$= ( \frac{b {}^{ - 3 + 7 - 2} }{b {}^{2 + 6} } ) {}^{ - 2}$
$= ( \frac{b {}^{2} }{b {}^{8} } ) {}^{ - 2}$
$= ( \frac{1}{b {}^{6} } ) {}^{ - 2}$
$= (b {}^{6} ) {}^{2}$
$= b {}^{12}$