Find $\large \left(\frac{2}{6}\right)^2 \cdot \left(\frac{9}{27}\right)^3$.

Question

Find $\large  \left(\frac{2}{6}\right)^2 \cdot \left(\frac{9}{27}\right)^3$.

ganeshie8 · Answer

$\huge  \left(\frac{2}{6}\right)^2 \cdot \left(\frac{9}{27}\right)^3$

anonymous · Answer

yep

ganeshie8 · Answer

look at the second fraction,
9 goes in 27 right ?

anonymous · Answer

yeah

ganeshie8 · Answer

how many times ?

anonymous · Answer

3

ganeshie8 · Answer

$\huge  \left(\frac{2}{6}\right)^2 \cdot \left(\frac{9}{9*3}\right)^3$


$\huge  \left(\frac{2}{6}\right)^2 \cdot \left(\frac{1}{3}\right)^3$

ganeshie8 · Answer

Oh look at first fraction,
2 also goes in 6, right ?

anonymous · Answer

yeah

ganeshie8 · Answer

$\huge  \left(\frac{2}{6}\right)^2 \cdot \left(\frac{9}{9*3}\right)^3$


$\huge  \left(\frac{1}{3}\right)^2 \cdot \left(\frac{1}{3}\right)^3$

ganeshie8 · Answer

that simplifies to :



$\huge  \frac{1}{3^2 \bullet  3^3}$

ganeshie8 · Answer

further simplifies to :

$\huge  \frac{1}{3^{2+3} }$