Is 27 - (a - b)^3 a Difference of Cubes in Precalculus?

[mathdad]

Precalculus by David Cohen, 3rd Edition
Chapter 1, Section 1.3.
Question 54.

Factor the expression.

27 - (a - b)^3

Must I expand (a - b)^3 or is this expression the difference of cubes?

 

[kaliprasad]

Precalculus by David Cohen, 3rd Edition
Chapter 1, Section 1.3.
Question 54.

Factor the expression.

27 - (a - b)^3

Must I expand (a - b)^3 or is this expression the difference of cubes?

difference of 2 cubes $3^3-(a-b)^3$

 

[mathdad]

27 - (a - b)^3

3^3 - (a - b)^3

The difference of cubes formula:

a^3 – b^3 = (a – b)(a^2 + ab + b^2)

I will let a = 3 and b = (a - b).

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

 

[kaliprasad]

27 - (a - b)^3

3^3 - (a - b)^3

The difference of cubes formula:

a^3 – b^3 = (a – b)(a^2 + ab + b^2)

I will let a = 3 and b = (a - b).

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

(3 - a + b)(9 + 3a - 3b + (a - b)(a + b))

(3 - a + b)(9 + 3a - 3b + a^2 -- b^2)

I cannot simplify anymore.

there are 2 issues here

a = 3 and b = (a - b) is bad notation .your are using a for 2 different purposes. you could use $x^3-y^3$ with $x=3$ and $y = a-b$

secondly $(a-b)^2$ you have changed to $(a-b)(a+b)$

 

[mathdad]

there are 2 issues here

a = 3 and b = (a - b) is bad notation .your are using a for 2 different purposes. you could use $x^3-y^3$ with $x=3$ and $y = a-b$

secondly $(a-b)^2$ you have changed to $(a-b)(a+b)$

In that case, can you work it out for me step by step?

 

[kaliprasad]

In that case, can you work it out for me step by step?

your steps are correct till you instead of expanding $(a-b)^2$ did $(a-b)(a+b) $ and I work it out here

$27-(a-b)^3$
= $3^3-(a-b)^3$
$= (3-(a-b))(3^2 + (a-b) + (a-b)^2)$
$= (3-a + b) ( 3^2 + a - b + a^2-2ab +b^2)$

 

[MarkFL]

your steps are correct till you instead of expanding $(a-b)^2$ did $(a-b)(a+b) $ and I work it out here

$27-(a-b)^3$
= $3^3-(a-b)^3$
$= (3-(a-b))(3 + (a-b) + (a-b)^2)$

This last step should be:

\(\displaystyle \left(3-(a-b)\right)\left(3^2+3(a-b)+(a-b)^2\right)\) :D

 

[mathdad]

This last step should be:

\(\displaystyle \left(3-(a-b)\right)\left(3^2+3(a-b)+(a-b)^2\right)\) :D

I made it that far and beyond.

 

[kaliprasad]

This last step should be:

\(\displaystyle \left(3-(a-b)\right)\left(3^2+3(a-b)+(a-b)^2\right)\) :D

Thanks mark. my mistake

 

[mathdad]

I had it right all along.