$$\sqrt[3]{-128}$$

Help me solve!

Question

$$\sqrt[3]{-128}$$

Help me solve!

anonymous · Answer

I would divide 128 by 2 and keep dividingit by 2 until u can no longer divideit by 2.  Then each triad of 2's is one factor of the cube root.  Collect all your triad of 2's.

anonymous · Answer

$$128$$ is a power of 2, it is $$128=2^7$$

anonymous · Answer

I multiplied 5 three times and ended up with 125

anonymous · Answer

How would I find $$\sqrt[3]{-128}$$

e.mccormick · Answer

First, when taking a nagative out of a root, that can happen in cubes.  Now, with a cube root, you have to take them out in sets of three.  As satellite73 pointed out, this is $2^7$ so you really have:
$\sqrt[3]{-(2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2)} $.  How many sets of three, 2s can you get out of that?

anonymous · Answer

2 sets.

e.mccormick · Answer

OK.  What would the two sets become outside the radical?

anonymous · Answer

4?

e.mccormick · Answer

Yes!  So now you have something with a 4 and a cube root of 2.  The question remainging is the negative sign.

e.mccormick · Answer

Becuase it is a cube root, what do you think happens with the negative?

anonymous · Answer

It becomes a -4?

e.mccormick · Answer

Yes.  However, the - inside the cube goes away.  $-4\sqrt[3]{2}$ is the best answer.   $4\sqrt[3]{-2}$ is an OK answer because it still keeps the negative, but not how it would usually be seen.    $-4\sqrt[3]{-2}$ is a BAD answer and wrong because it has two negatives and therefore is positive.

e.mccormick · Answer

And if you put   $-4\sqrt[3]{2}$ and   $\sqrt[3]{-128}$ in a calculator, you should get the same decimal aproximation answer for both!