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- anonymous

Can someone help me with this cube roots question?

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- anonymous

Can someone help me with this cube roots question?

- schrodinger

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- BloomLocke367

I can try :)

- anonymous

- anonymous

I know you have to substitute in the values, but I can't figure out how to simplify to get a whole number. We haven't done imaginaries yet so I know I can't use those.

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- BloomLocke367

What do you mean you can't figure out how to simplify to get a whole number?

- BloomLocke367

You know what a cubed number is, right?

- anonymous

All the values in the example questions are getting whole numbers for y. I don't know how when they aren't perfect cubes.

- anonymous

Yes

- BloomLocke367

They are perfect cubes >.<

- anonymous

I'm not sure what to do then

- BloomLocke367

And take a look at that first one you did. You made a small mistake.

- BloomLocke367

How did you get 1?

- anonymous

Oh wait is it -1?

- BloomLocke367

\[\sqrt[3]{-1}=-1\] and then you also have to subtract 2 from that.
\(-1-2=?\)

- anonymous

Oops, -3.

- anonymous

Then how do I find the cube root of -3?

- BloomLocke367

you don't need to find the cube root of -3. Now that you have -3, all that it means is that \[-3=\sqrt[3]{-1}-2\]

- BloomLocke367

so -3 is your y-value. The only thing you need to take the cube root of is the x-value that was given to you

- anonymous

So the next one would be -2?

- BloomLocke367

Yes! \(\Huge\color{lime}{\checkmark}\)

- anonymous

Then 0, then 2?

- BloomLocke367

Yep!

- anonymous

Thank you so much!!

- BloomLocke367

You're welcome! I'm glad I was able to help :)

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