How do you factor (p)^3 - 8c^3 when p = (a + b) and b = 2c?

[mathdad]

Factor (a + b)^3 - 8c^3.

Difference of cubes, right?

I say in the formula, a = (a + b) and b = 2c.

Right?

 

[Greg]

$$(a-b)^3=a^3-3a^2b+3ab^2-b^3$$

Rearrange:

$$\begin{align*}a^3-b^3&=(a-b)^3+3a^2b-3ab^2 \\
&=(a-b)^3+3ab(a-b) \\
&=(a-b)[(a-b)^2+3ab] \\
&=(a-b)(a^2-2ab+b^2+3ab) \\
&=(a-b)(a^2+ab+b^2)\end{align*}$$

$$(a+b)^3-8c^3=(a+b-2c)[(a+b)^2+2c(a+b)+4c^2]$$

 

[topsquark]

Factor (a + b)^3 - 8c^3.

Difference of cubes, right?

I say in the formula, a = (a + b) and b = 2c.

Right?

I mentioned this in another thread. Don't set the LHS of "a = a + b" It's too confusing.

-Dan

 

[mathdad]

I mentioned this in another thread. Don't set the LHS of "a = a + b" It's too confusing.

-Dan

I got it. I will let p = (a + b). I will then substitute into the difference of cubes formula and simplify as much as possible.