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

[mathdad]

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.

 

[kaliprasad]

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

 

[topsquark]

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

 

[mathdad]

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?

 

[MarkFL]

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:

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

 

[mathdad]

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!