How to express 80^2/3 as a product of powers of prime numbers

Question

So the given was: $(80)^\frac23(25)^\frac32$ and I was told to simplify and express as a product of powers of prime numbers.

Now I'm not very familiar with this product of powers of prime numbers so correct me if I'm wrong but what I did first was simplify $(25)^\frac32$ to $125$ and then (presumably) to express it as a product of powers of prime numbers, I divided it to a Prime number that was most likely a factor which was $5$ and ended up with $5\cdot5\cdot5$ or $5^3$ (again correct me if I'm wrong) so now we have $$(80)^\frac23\cdot5^3$$ we can't simplify $(80)^\frac23$ because $\sqrt[3]{6400}$ doesn't have a simple answer so what I did was express $80$ as a product of powers of prime numbers which became $2\cdot2\cdot2\cdot2\cdot5$ or simply $2^4\cdot5$ the final answer being: $$(2^4\cdot5)^\frac23\cdot5^3$$

Is this the simplest it can get? Did I do something wrong?

Answer 1

$$80^{2/3}\cdot25^{3/2}=(2^4\cdot5)^{2/3}\cdot(5^2)^{3/2}=(2^4)^{2/3}\cdot5^{2/3}\cdot5^3=2^{8/3}\cdot5^{11/3}$$

Answer 2

Weird way of putting it but:

$80$ can be expressed in terms of prime factorization as $2^4\cdot5$. And $25$ as $5^2$

So $(80)^{\frac 23}\cdot(25)^{\frac 32}= $

$(2^4\cdot 5)^{\frac 23}\cdot (5^2)^{\frac 32} = $

$(2^{4\cdot \frac 23}\cdot 5^{\frac 23})\cdot(5^{2\cdot\frac 32})=$

$2^{\frac 83}\cdot 5^{\frac 23}\cdot 5^3=$

$2^{\frac 83}\cdot 5^{\frac 23 + 3}=$

$2^{\frac 83}\cdot 5^{\frac {11}3}$

Now I have no idea why anyone would want to express it that way....

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With experience and comfort so things become clearer and apparent.

$25^{\frac 12} = 5$ and $8^{\frac 13} = 2$ so

$(80)^{\frac 23}\cdot (25)^{\frac 32} =$

$(2^2\cdot 10^{\frac 23})\cdot (5^3) = $

$2^2\cdot 2^{\frac 23}\cdot 5^{\frac 23}\cdot 5^3 = $

$2^{2 + \frac 23}5^{3 + \frac 23}=$

$2^{\frac 83}5^{\frac {11}3}$.

That's perhaps a little slicker but not as obvious or as illustrative to learn by.