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

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Discussion Overview

The discussion revolves around the factorization of the expression (p)^3 - 8c^3, specifically when p is defined as (a + b) and b is equal to 2c. Participants explore the application of the difference of cubes formula in this context.

Discussion Character

  • Mathematical reasoning, Technical explanation, Debate/contested

Main Points Raised

  • One participant suggests that the expression can be factored using the difference of cubes formula, proposing that a = (a + b) and b = 2c.
  • Another participant provides a detailed mathematical derivation of the difference of cubes, rearranging and simplifying the expression to show how it can be factored.
  • Some participants express confusion regarding the substitution of variables, specifically cautioning against setting the left-hand side of the equation as "a = a + b" due to potential confusion.
  • A later reply indicates a decision to let p = (a + b) and to substitute this into the difference of cubes formula for further simplification.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to factor the expression, with some expressing confusion and others providing differing methods of factorization.

Contextual Notes

There are unresolved issues regarding the clarity of variable substitution and the potential for confusion in the expressions used. Some mathematical steps remain unverified or unclear.

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

Difference of cubes, right?

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

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

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.
 

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