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

Question

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

Answer 1

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.

Answer 2

**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 $\displaystyle \frac 53 \cdot -\frac 35$?

Answer 3

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