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I need help in solving this problem (sorry I didn't know how to write it on here).

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closed as off-topic by Cookie, Hans Engler, Andrés E. Caicedo, M Turgeon, qwr Jun 17 '14 at 1:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Cookie, Hans Engler, Andrés E. Caicedo, M Turgeon, qwr
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ I think the question is evaluating $(5/3)^3(-3/5)^2$ into simplest terms. $\endgroup$ – Kaj Hansen Jun 17 '14 at 0:10
  • $\begingroup$ Usually an equation needs an equals sign. Do you mean $\frac 53 x^3 = - \left(\frac{3}{5}\right)^2$? $\endgroup$ – Sten Jun 17 '14 at 0:10
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    $\begingroup$ 404 calculus/linear algebra. $\endgroup$ – Shahar Jun 17 '14 at 0:10
  • $\begingroup$ Sorry It's not an equation $\endgroup$ – LuisDavis Jun 17 '14 at 0:18
  • $\begingroup$ Do you know what powers are? Do you know how to multiply rational numbers together? Do you know that negative times negative is positive? If the answer to all three of these questions is yes, then you have all the tools you need to to simplify the expression. All you have to do is use them. $\endgroup$ – blue Jun 17 '14 at 0:22
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Hint:

$\left(\dfrac{5}{3}\right)^3*\left(-\dfrac{3}{5}\right)^2$ = $\left(\dfrac{5^3}{3^3}\right)*\left(\dfrac{(-3)^2}{5^2}\right)$

Another hint:

To multiply two fractions, you can multiply the two numerators by each other, and the two denominators by each other to get the new numerator and denominator.

Here is a resource you should work through.

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**Hint:**$$\left(\frac 53 \right)^3 \left(-\frac 35 \right)^2=\left(\frac 53 \right)\left(\frac 53 \right)^2 \left(-\frac 35 \right)^2=\left(\frac 53 \right)\left(\frac 53 \cdot-\frac 35 \right)^2$$ Can you simplify $\displaystyle \frac 53 \cdot -\frac 35$?

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$\begin{align}\left(\dfrac{5}{3}\right)^3\left(\dfrac{-3}{5}\right)^2 & = \left(\dfrac{5^3}{3^3}\right)\cdot\left(\dfrac{(-1)^2 3^2}{5^2}\right) & \text{by commutativity of exponents} \\ ~ & = \dfrac{(-1)^2 5^1}{3^1} & \text{by associativity of exponents} \\ ~ & = \dfrac{5}{3} & \text{by }(-1)^2=1, a^1 = a \end{align}$

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