Help with alg prob

**lax600** · May 12th 2008, 05:51 PM

$x^3 + x^2 - 8y^3 - 4y^2$

i tried doing

$=[x + (-2y)]^3 + x^2 - 4y^2 <br /> =[x + (-2y)][(x)^2 - (x)(-2y) + (-2y)^2) + x^2 - 4y^2$
$= [x + (-2y)][(x)^2 -(x)(-2y) + (-2y)^2]$

but I'm pretty sure thats not right...

**Reckoner** · May 12th 2008, 07:04 PM

Hi, lax600! I assume you are being asked to factor this expression?

Originally Posted by lax600

$x^3 + x^2 - 8y^3 - 4y^2$

i tried doing

$=[x + (-2y)]^3 + x^2 - 4y^2 <br /> =[x + (-2y)][(x)^2 - (x)(-2y) + (-2y)^2) + x^2 - 4y^2$
$= [x + (-2y)][(x)^2 -(x)(-2y) + (-2y)^2]$

but I'm pretty sure thats not right...

This is indeed wrong. Be sure to remember: you cannot distribute exponents over a sum! That is, $\left(a + b\right)^n\neq a^n + b^n$ in general. In this case, $x^3 - \left(2y\right)^3\neq\left(x - 2y\right)^3$ .

Instead, you should observe that your expression contains both a difference of cubes and a difference squares, which can be factored through their simple formulas, and then you will be left with two terms that contain a common factor.

First, let's group things a bit differently:

$x^3 + x^2 - 8y^3 - 4y^2$

$=\left(x^3 - 8y^3\right) + \left(x^2 - 4y^2\right)$

Now, you should know how to factor differences of squares and sums or differences of cubes:

$a^2 - b^2 = \left(a + b\right)\left(a - b\right)$

$a^3 \pm b^3 = \left(a \pm b\right)\left(a^2 \mp ab + b^2\right)$

So, using these properties we get:

$\left(x^3 - 8y^3\right) + \left(x^2 - 4y^2\right)$

$=\left(x - 2y\right)\left(x^2 + 2xy + 4y^2\right) + \left(x - 2y\right)\left(x + 2y\right)$

Can you take it from here?

Edit: You actually seemed to have been doing exactly as I have just explained, but I was thrown off by your bad notation. The original work you posted is mostly okay, but the first line should be written with the two terms cubed separately, and then in the second line your $x^2 - 4y^2$ just seems to have disappeared.

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