How to solve $\left(\dfrac{5}{3}\right)^3\left(-\dfrac{3}{5}\right)^2$ [closed]

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

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

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• I think the question is evaluating $(5/3)^3(-3/5)^2$ into simplest terms. – Kaj Hansen Jun 17 '14 at 0:10
• Usually an equation needs an equals sign. Do you mean $\frac 53 x^3 = - \left(\frac{3}{5}\right)^2$? – Sten Jun 17 '14 at 0:10
• 404 calculus/linear algebra. – Shahar Jun 17 '14 at 0:10
• Sorry It's not an equation – LuisDavis Jun 17 '14 at 0:18
• 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. – blue Jun 17 '14 at 0:22

$\left(\dfrac{5}{3}\right)^3*\left(-\dfrac{3}{5}\right)^2$ = $\left(\dfrac{5^3}{3^3}\right)*\left(\dfrac{(-3)^2}{5^2}\right)$
**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 $\displaystyle \frac 53 \cdot -\frac 35$?
\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}