887/VB02. problem / Math - Solved Math Problems

MR-887 / 14b. problem

Simplify the following expression:

$({(\frac{3 a^{2}}{4 b^{- 3}})}^{- 3} \div {(\frac{9 a^{- 2} b}{4})}^{- 2}) \cdot \frac{b^{7}}{12 a^{- 11}}$

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Source: Vene T. Bogoslavov: Collection of Solved Problems #2 - 14.b.)

Algebra → Exponents → Powers

$({(\frac{3 a^{2}}{4 b^{- 3}})}^{- 3} \div {(\frac{9 a^{- 2} b}{4})}^{- 2}) \cdot \frac{b^{7}}{12 a^{- 11}}$

$= ({(\frac{3 a^{2} b^{3}}{4})}^{- 3} \div {(\frac{9 b}{4 a^{2}})}^{- 2}) \cdot \frac{b^{7} \cdot a^{11}}{12}$

$= ({(\frac{4}{3 a^{2} b^{3}})}^{3} \div {(\frac{4 a^{2}}{9 b})}^{2}) \cdot \frac{b^{7} \cdot a^{11}}{12}$

$= ((\frac{4^{3}}{3^{3} \cdot a^{2 \cdot 3} \cdot b^{3 \cdot 3}}) \div (\frac{4^{2} \cdot a^{2 \cdot 2}}{9^{2} \cdot b^{2}})) \cdot \frac{a^{11} \cdot b^{7}}{12}$

$= ((\frac{64}{81 \cdot a^{6} \cdot b^{9}}) \div (\frac{16 \cdot a^{4}}{81 \cdot b^{2}})) \frac{a^{11} \cdot b^{7}}{12}$

$= (\frac{64}{27 \cdot a^{6} \cdot b^{9}} \cdot \frac{81 \cdot b^{2}}{16 \cdot a^{4}}) \cdot \frac{a^{11} \cdot b^{7}}{12}$

$= (\frac{4}{a^{6} b^{7}} \cdot \frac{3}{a^{4}}) \cdot \frac{a^{11} \cdot b^{7}}{12}$

$= \frac{12}{a^{10} \cdot b^{7}} \cdot \frac{a^{11} \cdot b^{7}}{12}$

$= a$

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