Question

(-2x²y)³ / (xy²z)² * (-yz)² Hello guys, im having trouble with this babe...

at first I thought it could be equal to:
= [(-2 x² y)² * (-2 x² y)] * (-2x² y)¹ / [(x y² z) * (x y² z)] * [(-y z) * (-y z)]

= +4x² y² * (-2 x² y) / x² y^6 z^4

= -8x^6 y³ / x² y^6 z^4 , and from here I don't know how to proceed...

Can somebody tell me where I'm going wrong, please?

Answer 1

$\frac{(-2x^2y)^3}{(xy^2z)^2(-yz)^2}$

On top: (-2)³ = -2 × -2 × -2 = -8
(x²)³ = x^6 (the exponents multiply)
and of course, (y)³ = y³

On the bottom: (xy²z)² = x² y^4 z²
(-yz)² = y²z²
Multiplying these together, the exponents add and we get x² y^6 z^4.

$\frac{-8x^6y^3}{x^2y^6z^4}$

So, your reasoning is correct for what you have so far.

Your next step would be cancelling shared factors from the top and bottom.
Just like with regular fractions, if the numerator and denominator are divisible by the same number, you can divide them by it to simplify. (ex: 4/6 = 2/3)

Well, x^6 and x^2 are both divisible by x^2, right?
We can also cancel the y^3.

It might help to visualise the factors like this:

$\frac{-8xxxxxxyyy}{xxyyyyyyzzzz}$

Once you've cancelled out x² and y³ from each, you're left with

$\boxed{\frac{-8x^4}{y^3z^4}}$