simplify

Question

simplify

e.mccormick · Answer

For a cube root, you need to pull out in sets of three.

e.mccormick · Answer

$\sqrt[3]{a^4}\implies \sqrt[3]{a (a\cdot a\cdot a)}$ set of three of the variable a, so:

$a\sqrt[3]{a}$

anonymous · Answer

would the denominator be 2?

e.mccormick · Answer

Huh?  There is no denominator in what you linked.

anonymous · Answer

oh?

anonymous · Answer

attachment

e.mccormick · Answer

Ah.  Different problem.  OK.  Know how fraction division is multiplicaton of the inverse?

anonymous · Answer

kind of

e.mccormick · Answer

It just means we can do this:$$\frac{
\frac{x^2+6x+9}{8x}
}{
\frac{x+3}{4x}
}
=
\frac{x^2+6x+9}{8x}\cdot \frac{4x}{x+3}
$$

e.mccormick · Answer

Now, in that form on the right, if you factor the top of the one fraction, I bet it will cancel with the bottom of the other.  And, I see some canceling between the bottom left and the upper right!  Because these are multiplied fractions, that is valid.

e.mccormick · Answer

And, to answer your other question, the denominator does go to 2.

anonymous · Answer

so |dw:1372133961181:dw|

anonymous · Answer

or would it be x +3 ?

e.mccormick · Answer

Hehe... you caught hwat I was going to ask about.  It is x+3, not -.

e.mccormick · Answer

$x^2+6x+9$ factors to $(x+3)^2$  FOIL it if you are not sure.