Solution 1.3:5f

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The whole expression is quite complicated, so it can be useful to simplify the terms
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The whole expression is quite complicated, so it can be useful to simplify the terms <math>\bigl(125^{\frac{1}{3}}\bigr)^{2}</math> and <math>\bigl(27^{\frac{1}{3}}\bigr)^{-2}</math> first,
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<math>\left( 125^{\frac{1}{3}} \right)^{2}</math>
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and
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<math>\left( 27^{\frac{1}{3}} \right)^{-2}</math>
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first:
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{{Displayed math||<math>\begin{align}
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\bigl(125^{\frac{1}{3}}\bigr)^{2} &= 125^{\frac{1}{3}\cdot 2} = 125^{\frac{2}{3}}\,,\\[5pt]
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\bigl(27^{\frac{1}{3}}\bigr)^{-2} &= 27^{\frac{1}{3}\cdot (-2)} = 27^{-\frac{2}{3}}\,\textrm{.}\end{align}</math>}}
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<math>\left( 125^{\frac{1}{3}} \right)^{2}=125^{\frac{1}{3}\centerdot 2}=125^{\frac{2}{3}}</math>
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Then, the bases 125, 27 and 9 can be rewritten as
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<math>\left( 27^{\frac{1}{3}} \right)^{-2}=27^{\frac{1}{3}\centerdot \left( -2 \right)}=27^{-\frac{2}{3}}</math>
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Then, the bases
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<math>125,\ \ 27</math>
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and
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<math>9</math>
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can be rewritten as
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<math>\begin{align}
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& 125=5\centerdot 25=5\centerdot 5\centerdot 5=5^{3} \\
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& \\
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& 27=3\centerdot 9=3\centerdot 3\centerdot 3=3^{3} \\
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& \\
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& 9=3\centerdot 3=3^{2} \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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125 &= 5\cdot 25 = 5\cdot 5\cdot 5 = 5^{3},\\
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27 &= 3\cdot 9 = 3\cdot 3\cdot 3 = 3^{3},\\
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9 &= 3\cdot 3 = 3^{2}\textrm{.}
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\end{align}</math>}}
With the help of the power rules,
With the help of the power rules,
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\bigl(125^{\frac{1}{3}}\bigr)^{2}\cdot\bigl(27^{\frac{1}{3}}\bigr)^{-2}\cdot 9^{\frac{1}{2}} &= 125^{\frac{2}{3}}\cdot 27^{-\frac{2}{3}}\cdot 9^{\frac{1}{2}}\\[5pt]
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& \left( 125^{\frac{1}{3}} \right)^{2}\centerdot \left( 27^{\frac{1}{3}} \right)^{-2}\centerdot 9^{\frac{1}{2}}=125^{\frac{2}{3}}\centerdot 27^{-\frac{2}{3}}\centerdot 9^{\frac{1}{2}} \\
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&= \bigl(5^{3}\bigr)^{\frac{2}{3}}\cdot \bigl(3^{3}\bigr)^{-\frac{2}{3}}\cdot \bigl(3^{2}\bigr)^{\frac{1}{2}}\\[5pt]
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& \\
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&= 5^{3\cdot\frac{2}{3}}\cdot 3^{3\cdot (-\frac{2}{3})}\cdot 3^{2\cdot\frac{1}{2}}\\[5pt]
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& =\left( 5^{3} \right)^{\frac{2}{3}}\centerdot \left( 3^{3} \right)^{-\frac{2}{3}}\centerdot \left( 3^{2} \right)^{\frac{1}{2}}=5^{3\centerdot \frac{2}{3}}\centerdot 3^{3\centerdot \left( -\frac{2}{3} \right)}\centerdot 3^{2\centerdot \frac{1}{2}} \\
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&= 5^{2}\cdot 3^{-2}\cdot 3^{1}\\[5pt]
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& \\
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&= 5^{2}\cdot 3^{-2+1}\\[5pt]
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& =5^{2}\centerdot 3^{-2}\centerdot 3^{1}=5^{2}\centerdot 3^{-2+1}=5^{2}\centerdot 3^{-1}=5\centerdot 5\centerdot \frac{1}{3} \\
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&= 5^{2}\cdot 3^{-1}\\[5pt]
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& \\
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&= 5\cdot 5\cdot \frac{1}{3}\\[5pt]
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& \frac{25}{3} \\
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&= \frac{25}{3}\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}

Current revision

The whole expression is quite complicated, so it can be useful to simplify the terms \displaystyle \bigl(125^{\frac{1}{3}}\bigr)^{2} and \displaystyle \bigl(27^{\frac{1}{3}}\bigr)^{-2} first,

\displaystyle \begin{align}

\bigl(125^{\frac{1}{3}}\bigr)^{2} &= 125^{\frac{1}{3}\cdot 2} = 125^{\frac{2}{3}}\,,\\[5pt] \bigl(27^{\frac{1}{3}}\bigr)^{-2} &= 27^{\frac{1}{3}\cdot (-2)} = 27^{-\frac{2}{3}}\,\textrm{.}\end{align}

Then, the bases 125, 27 and 9 can be rewritten as

\displaystyle \begin{align}

125 &= 5\cdot 25 = 5\cdot 5\cdot 5 = 5^{3},\\ 27 &= 3\cdot 9 = 3\cdot 3\cdot 3 = 3^{3},\\ 9 &= 3\cdot 3 = 3^{2}\textrm{.} \end{align}

With the help of the power rules,

\displaystyle \begin{align}

\bigl(125^{\frac{1}{3}}\bigr)^{2}\cdot\bigl(27^{\frac{1}{3}}\bigr)^{-2}\cdot 9^{\frac{1}{2}} &= 125^{\frac{2}{3}}\cdot 27^{-\frac{2}{3}}\cdot 9^{\frac{1}{2}}\\[5pt] &= \bigl(5^{3}\bigr)^{\frac{2}{3}}\cdot \bigl(3^{3}\bigr)^{-\frac{2}{3}}\cdot \bigl(3^{2}\bigr)^{\frac{1}{2}}\\[5pt] &= 5^{3\cdot\frac{2}{3}}\cdot 3^{3\cdot (-\frac{2}{3})}\cdot 3^{2\cdot\frac{1}{2}}\\[5pt] &= 5^{2}\cdot 3^{-2}\cdot 3^{1}\\[5pt] &= 5^{2}\cdot 3^{-2+1}\\[5pt] &= 5^{2}\cdot 3^{-1}\\[5pt] &= 5\cdot 5\cdot \frac{1}{3}\\[5pt] &= \frac{25}{3}\,\textrm{.} \end{align}

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