How do you simplify ${(\frac{8}{27})}^{- \frac{2}{3}}$ $(8/27)^(-2/3)$ ?

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How do you simplify ${(\frac{8}{27})}^{- \frac{2}{3}}$ $(8/27)^(-2/3)$ ?

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Answer 1 · 2017-02-14T21:09:43

$\frac{9}{4}$ $9/4$

Explanation:

${(\frac{8}{27})}^{- \frac{2}{3}}$ $(8/27)^(-2/3)$
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$= {(\frac{27}{8})}^{\frac{2}{3}}$ $=(27/8)^(2/3)$
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The prime factorization of 27 and 8 is:
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$27 = 9 \times 3 = 3 \times 3 \times 3 = 3^{3}$ $27=9xx3=3xx3xx3=3^3$
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$8 = 4 \times 2 = 2 \times 2 \times 2 = 2^{3}$ $8=4xx2=2xx2xx2=2^3$

Substituting the factorization on the above fraction we have:
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${(\frac{27}{8})}^{\frac{2}{3}}$ $(27/8)^(2/3)$
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$= {(\frac{3^{3}}{2^{3}})}^{\frac{2}{3}}$ $=(3^3/2^3)^(2/3)$
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$= {({(\frac{3}{2})}^{3})}^{\frac{2}{3}}$ $=((3/2)^3)^(2/3)$
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Applying the property of power of a power that says:
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${(a^{n})}^{m} = a^{m \times n}$ $color(red)((a^n)^m=a^(mxxn))$
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${({(\frac{3}{2})}^{3})}^{\frac{2}{3}}$ $((3/2)^3)^(2/3)$
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$= {(\frac{3}{2})}^{3 \times (\frac{2}{3})}$ $=(3/2)^(3xx(2/3))$
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$= {(\frac{3}{2})}^{2}$ $=(3/2)^2$
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$= \frac{9}{4}$ $=9/4$

Answer 2 · 2017-02-14T21:11:28

$= \frac{9}{4}$ $=9/4$

Explanation:

There are 3 different processes indicated in this expression.

Laws of indices:

$x^{- m} = \frac{1}{x^{m}} and {(\frac{a}{b})}^{- m} = {(\frac{b}{a})}^{m}$ $x^-m = 1/x^m" "and" "(a/b)^-m = (b/a)^m$

The second law is the one we will apply.

Also $x^{\frac{p}{q}} = {\sqrt[q]{x}}^{p}$ $x^(p/q) = rootq(x)^p$

${(\frac{8}{27})}^{- \frac{2}{3}} = {(\frac{27}{8})}^{+ \frac{2}{3}}$ $(8/27)^(-2/3) = (27/8)^(+2/3)$

$= {\sqrt[3]{\frac{27}{8}}}^{2} \leftarrow$ $= root3(27/8)^2" "larr$ find the cube roots first

$= {(\frac{3}{2})}^{2}$ $=(3/2)^2$

$= \frac{9}{4}$ $=9/4$

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