MHB Can I Use the Difference of Cubes Formula for Factoring 27 - (a - b)^3?

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The difference of cubes formula can be applied to factor the expression 27 - (a - b)^3 by identifying a as 3 and letting p equal (a - b). The formula states that a^3 - b^3 can be factored as (a - b)(a^2 + ab + b^2). After substituting the values, the expression simplifies to (3 - (a - b))(3^2 + 3(a - b) + (a - b)^2). Clarifying variable names is essential to avoid confusion in the calculations. This approach effectively demonstrates the application of the difference of cubes formula.
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Factor 27 - (a - b)^3.

Can I apply the difference of cubes formula here?

If this is true, then in the formula a = 3 and b = (a - b).

Right?

Note: 27 became 3^3 leading to a = 3 in the difference of cubes formula.
 
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RTCNTC said:
Factor 27 - (a - b)^3.

Can I apply the difference of cubes formula here?

If this is true, then in the formula a = 3 and b = (a - b).

Right?

Note: 27 became 3^3 leading to a = 3 in the difference of cubes formula.

Yes.

Then in case of doubt you after factoring can multiply and see the result
 
RTCNTC said:
If this is true, then in the formula a = 3 and b = (a - b).
You need to clarify your variables. "b = (a - b)" isn't right. Pick a variable name, say, p... Anything but that b on the LHS... Then you have p = a - b.

-Dan
 
topsquark said:
You need to clarify your variables. "b = (a - b)" isn't right. Pick a variable name, say, p... Anything but that b on the LHS... Then you have p = a - b.

-Dan

In the difference of cubes, let a = 3 and p = (a - b).

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

a^3 - b^3 = (3 - p)(3^3 + 3p + p^2)

I must now simplify The RHS and then back-substitute for p.

Yes?
 
It might be more clear to state something like the following:

The difference of cubes formula states:

$$p^3-q^3=(p-q)\left(p^2+pq+q^2\right)$$

And so, given:

$$3^3-(a-b)^3$$

If we let $p=3$ and $q=a-b$, we obtain:

$$3^3-(a-b)^3=(3-(a-b))\left(3^2+3(a-b)+(a-b)^2)\right)$$
 
MarkFL said:
It might be more clear to state something like the following:

The difference of cubes formula states:

$$p^3-q^3=(p-q)\left(p^2+pq+q^2\right)$$

And so, given:

$$3^3-(a-b)^3$$

If we let $p=3$ and $q=a-b$, we obtain:

$$3^3-(a-b)^3=(3-(a-b))\left(3^2+3(a-b)+(a-b)^2)\right)$$

Great substitution!
 
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