How do you divide $\frac{{(a^{- 4} b^{- 1} c^{3})}^{- 3}}{{(a^{- 2} b^{5} c^{- 4})}^{- 4}}$ $\frac { ( a ^ { - 4} b ^ { - 1} c ^ { 3} ) ^ { - 3} } { ( a ^ { - 2} b ^ { 5} c ^ { - 4} ) ^ { - 4} }$ ?

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Answer 1 · 2018-07-06T17:59:35

$\frac{a^{4} \cdot b^{23}}{c^{25}}$ $(a^4*b^23)/c^25$

We will use that

${(a^{n} b^{m})}^{p} = a^{n p} b^{m p}$ $(a^nb^m)^p=a^(np)b^(mp)$
and

$\frac{a^{n}}{b^{m}} = a^{n - m}$ $(a^n)/b^m=a^(n-m)$
so we get

$\frac{a^{12} \cdot b^{3} \cdot c^{- 9}}{a^{8} \cdot b^{- 20} \cdot c^{16}}$ $(a^12*b^3*c^(-9))/(a^8*b^(-20)*c^16)$
simplifying further

$\frac{a^{4} \cdot b^{23}}{c^{25}}$ $(a^4*b^23)/c^25$

How do you divide \frac { ( a ^ { - 4} b ^ { - 1} c ^ { 3} ) ^ { - 3} } { ( a ^ { - 2} b ^ { 5} c ^ { - 4} ) ^ { - 4} }(a−4b−1c3)−3(a−2b5c−4)−4\frac { ( a ^ { - 4} b ^ { - 1} c ^ { 3} ) ^ { - 3} } { ( a ^ { - 2} b ^ { 5} c ^ { - 4} ) ^ { - 4} }?