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

  • Thread starter Thread starter mathdad
  • Start date Start date
AI Thread Summary
To factor (p)^3 - 8c^3 where p = (a + b) and b = 2c, the difference of cubes formula is applied. The expression simplifies to (a + b - 2c)[(a + b)^2 + 2c(a + b) + 4c^2]. The formula used for the difference of cubes is a^3 - b^3 = (a - b)(a^2 + ab + b^2). The discussion emphasizes the importance of correctly substituting values into the formula without confusion. Ultimately, the goal is to simplify the expression effectively using the difference of cubes method.
mathdad
Messages
1,280
Reaction score
0
Factor (a + b)^3 - 8c^3.

Difference of cubes, right?

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

Right?
 
Mathematics news on Phys.org
$$(a-b)^3=a^3-3a^2b+3ab^2-b^3$$

Rearrange:

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

$$(a+b)^3-8c^3=(a+b-2c)[(a+b)^2+2c(a+b)+4c^2]$$
 
RTCNTC said:
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
 
topsquark said:
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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top