Question

**MEDAL FOR BEST ANSWER!!** @mathstudent55

Answer 1

Factor this expression completely, then place the factors in the proper location on the grid.

x^6 - y^6

Answer 2

Notice that the exponents are both 6.
Is x^6 the cube of something?
Is y^6 the cube of something?

Answer 3

I have no idea what it is the cube of.

Answer 4

Remember this rule:

\((a^m)^n = a^{mn} \)

Example:

\((x^4)^3 = x^{12} \)

Answer 5

So it's the cube of 2?

Answer 6

Now think of \(x^6\).

When you raise a power to a power, you multiply powers, so since 6, the exponent , is equal to 2 * 3, \(x^6 = (x^2)^3\).

Answer 7

That means x^6 is the cube of x^2.

Answer 8

Okay so x^2-y^2. . . ?

Answer 9

You are on the right track, but I need to change a little the steps.

Let's factor it it this way first.

Just like \(x^6 = (x^2)^3\)

\(x^6 = (x^3)^2 \), right?

Answer 10

Yeah. So how do I put that in the equation? I have a graph that looks like: _x_____-_)_x________________^2_ and I need to figure out what goes in the blanks (which are the _)

Answer 11

Let's rewrite x^6 as the square of something instead of as the cube of something.

Answer 12

\(x^6 - y^6 = (x^3)^2 - (y^3)^2 \)

Now let's factor the difference of two squares first.

\( (x^3)^2 - (y^3)^2 \)

\(= (x^3 + y^3)(x^3 - y^3) \)

Answer 13

Now we need to factor the sum of cubes and the difference of cubes.

Answer 14

How do I do that?

Answer 15

\(= (x^3 + y^3)(x^3 - y^3)\)

\( (x + y)(x^2 - xy + y^2)(x - y)(x ^2 + xy + y^2 )\)

Answer 16

That doesn't fit in my spaces though. It's _x_____-_)_x________________^2_

Answer 17

That's the complete factoring.

Answer 18

Okay so how do I get to the complete factoring?

Answer 19

What I did is the complete factoring of the expression x^6 - y^6.
That's what the problem asked for.

Answer 20

That doesnt fit in the spaces though.

Answer 21

Since your earlier problem today was about factoring the difference of 2 cubes, maybe all this problem is looking for is for it to be factored as the difference of two cubes. If so, then you do this:
\(x^6 - y^6 = (x^2)^3 - (y^2)^3 = (x^2 - y^2)(x^4 + x^2y^2 + y^4)\)

The problem here is that it is not completely factored as the instructions ask to do.

Answer 22

Here's an example of a difference of cubes:

\(8y^3 - 27 \)

\(= (2y)^3 - 3^3 \)

\( =(2y - 3)[ (2y)^2 + 2y \cdot 3 + 3^2\)]

\( = (2y - 3)(4y^2 + 6y + 9) \)

Answer 23

Let me send you a picture. Hold on. @mathstudent55

Answer 24

I need the answer to fit in those boxes.