Thread: Simplifying

Hello,

Not sure if I've simplified this as far as I can - pretty sure I can go further, just not sure what I should be doing (factoring, distributive)

Equation is (z + 1)^2 (3xy^2)^3 / (9y^2)^(1/2) x^(-3) (z+1)^(-1)

I've simplified it to...

(z^2 + 1) (27x^3y^6) / 3y (1/x^3) (1/z+1)

What should I be looking at next? Finding a common factor between both numerator and denominator (I can see three, along with each of the variables). I'd assume before I can factor out though, I should be expanding the brackets further - is that possible at this stage, and if so, how do I go about doing the denominator.

As you can see, I'm pretty lost as to what I should be doing next, so any tips would be great!

Originally Posted by astuart

Hello,

Not sure if I've simplified this as far as I can - pretty sure I can go further, just not sure what I should be doing (factoring, distributive)

Equation is (z + 1)^2 (3xy^2)^3 / (9y^2)^(1/2) x^(-3) (z+1)^(-1)

I've simplified it to...

(z^2 + 1) (27x^3y^6) / 3y (1/x^3) (1/z+1)

What should I be looking at next? Finding a common factor between both numerator and denominator (I can see three, along with each of the variables). I'd assume before I can factor out though, I should be expanding the brackets further - is that possible at this stage, and if so, how do I go about doing the denominator.

As you can see, I'm pretty lost as to what I should be doing next, so any tips would be great!

...

Originally Posted by Prove It

...

Gah, )z+1)^2 would equal z^2 + 2z + 1 or something along those lines, right?

I think i'll tackle this in the morning, not at 1 in the morning.

Also , it's

Originally Posted by Prove It

Also , it's

Right, I thought that was the case, but I just wasn't sure how to use the equation tags properly.

So I have...


Not sure if I an simplify between the numerator and denominator. There's a common factor of three there. Can I cancel out the z and the x? I didn't think I could because the z and the x are underneath a fraction (As in underneath a 1).