How do you simplify $\frac{(16 b^{- 2} c) (25 b^{4} c^{- 5})}{(15 b^{5} c^{- 1}) (8 b^{- 7} c^{- 2})}$ $((16b^-2c)(25b^4c^-5))/((15b^5c^-1)(8b^-7c^-2))$ ?

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How do you simplify $\frac{(16 b^{- 2} c) (25 b^{4} c^{- 5})}{(15 b^{5} c^{- 1}) (8 b^{- 7} c^{- 2})}$ $((16b^-2c)(25b^4c^-5))/((15b^5c^-1)(8b^-7c^-2))$ ?

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Answer 1 · 2016-06-23T10:49:12

$\frac{(16 b^{- 2} c) (25 b^{4} c^{- 5})}{(15 b^{5} c^{- 1}) (8 b^{- 7} c^{- 2})} = \frac{10 b^{4}}{3 c}$ $((16b^(-2)c)(25b^4c^(-5)))/((15b^5c^(-1))(8b^(-7)c^(-2)))=(10b^4)/(3c)$

Explanation:

For simplifying $\frac{(16 b^{- 2} c) (25 b^{4} c^{- 5})}{(15 b^{5} c^{- 1}) (8 b^{- 7} c^{- 2})}$ $((16b^(-2)c)(25b^4c^(-5)))/((15b^5c^(-1))(8b^(-7)c^(-2))$

we will use identities $a^{m} \times a^{n} = a^{m + n}$ $a^mxxa^n=a^(m+n)$ , $\frac{a^{m}}{a^{n}} = a^{m - n}$ $a^m/a^n=a^(m-n)$ , $a^{- m} = \frac{1}{a^{m}}$ $a^(-m)=1/a^m$ and $\frac{1}{a^{- m}} = a^{m}$ $1/a^(-m)=a^m$

Hence $\frac{(16 b^{- 2} c) (25 b^{4} c^{- 5})}{(15 b^{5} c^{- 1}) (8 b^{- 7} c^{- 2})}$ $((16b^(-2)c)(25b^4c^(-5)))/((15b^5c^(-1))(8b^(-7)c^(-2))$

= $\frac{(16 c) (25 b^{4}) b^{7} c^{2} \cdot c}{b^{2} c^{5} (15 b^{5}) (8)}$ $((16c)(25b^4)b^7c^2*c)/(b^2c^5(15b^5)(8)$

= $\frac{16 \times 25}{15 \times 8} \times \frac{c^{1} b^{4} b^{7} c^{2} \cdot c^{1}}{b^{2} c^{5} (b^{5})}$ $(16xx25)/(15xx8)xx(c^1b^4b^7c^2*c^1)/(b^2c^5(b^5)$

= $\frac{2 \times 2 \times 2 \times 2 \times 5 \times 5}{3 \times 5 \times 2 \times 2 \times 2} \times \frac{c^{(1 + 2 + 1)} b^{(4 + 7)}}{b^{(2 + 5)} c^{5}}$ $(2xx2xx2xx2xx5xx5)/(3xx5xx2xx2xx2)xx(c^((1+2+1))b^((4+7)))/(b^((2+5))c^5)$

= $\frac{2 \times 5}{3} \times \frac{c^{4} b^{11}}{b^{7} c^{5}} = \frac{10}{3} \times \frac{b^{(11 - 7)}}{c^{(5 - 4)}}$ $(2xx5)/(3)xx(c^4b^11)/(b^7c^5)=10/3xxb^((11-7))/c^((5-4))$

= $\frac{10 b^{4}}{3 c}$ $(10b^4)/(3c)$

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How do you simplify $\frac{(16 b^{- 2} c) (25 b^{4} c^{- 5})}{(15 b^{5} c^{- 1}) (8 b^{- 7} c^{- 2})}$ $((16b^-2c)(25b^4c^-5))/((15b^5c^-1)(8b^-7c^-2))$ ?

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How do you simplify (16b−2c)(25b4c−5)(15b5c−1)(8b−7c−2)((16b^-2c)(25b^4c^-5))/((15b^5c^-1)(8b^-7c^-2))?

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How do you simplify $\frac{(16 b^{- 2} c) (25 b^{4} c^{- 5})}{(15 b^{5} c^{- 1}) (8 b^{- 7} c^{- 2})}$ $((16b^-2c)(25b^4c^-5))/((15b^5c^-1)(8b^-7c^-2))$ ?