Hello, I cannot understand the circled stage of this equation.
Could someone please help explain how it occurred ?
Thanks kindly for any help.
They need to get everything to the smallest possible base. Notice that $\displaystyle \displaystyle 4 = 2^2$.
So $\displaystyle \displaystyle \frac{4^3 \times 5^2 \times 2^3}{3^3 \times 3^2 \times 5^3} = \frac{(2^2)^3 \times 5^2 \times 2^3}{3^3 \times 3^2 \times 5^3}$.
The next step is to cancel as many of the $\displaystyle \displaystyle 5$s as possible. Since there are $\displaystyle \displaystyle 3$ on the bottom and $\displaystyle \displaystyle 2$ on the top, that means you can cancel $\displaystyle \displaystyle 2$ of them and be left with $\displaystyle \displaystyle 3-2 = 1$ of them on the bottom.
Does it make a bit more sense now?
Numerator
As $\displaystyle 4 = 2^2$ you can write $\displaystyle 4^3 = (2^2)^3$
Hence the numerator becomes $\displaystyle (2^2)^3 \times 2^3 = 2^9$
Denominator:
As $\displaystyle a^{b+c} = a^ba^c$ it can be rewritten as $\displaystyle 3^3 \times 3^2 = 3^{3+2}$
$\displaystyle 5^{3-2}$ comes from $\displaystyle 5^2$ in the numerator so in the denominator it may be written as $\displaystyle 5^{-2}$ and put in the denominator. Combined with $\displaystyle 5^3$ we can put $\displaystyle 5^{3-2}$ on the denominator
In total we get
$\displaystyle \dfrac{(2^2)^3 \times 2^3}{3^{3+2} \times 5^{3-2}}$ which is the question.
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Personally though I'd have cancelled out 5^2 from top and bottom thus eliminating the need to deal with powers of 5.
Just look at the terms with 5 in you have $\displaystyle \dfrac{5^2}{5^3}$. You can write that at $\displaystyle 5^{2-3}$ if you like - as long as you leave it in the numerator.
If you want to look at the denominator (which is what the question does) then you will have to take the reciprocal of $\displaystyle 5^{2-3}$ which is $\displaystyle 5^{3-2}$ and put this term on the denominator.
If you can use your own working you'll save yourself a lot of time by saying that $\displaystyle \dfrac{5^2}{5^3} = \dfrac{1}{5}$