The discussion revolves around the expression 27 - (a - b)^3 and whether it can be factored as a difference of cubes in the context of precalculus. Participants explore the application of the difference of cubes formula and the implications of notation and expansion in the factoring process.
Participants express differing opinions on the notation and the steps involved in the factorization process. There is no consensus on the correctness of the various approaches and expansions discussed.
Participants highlight issues with notation and the potential for confusion in the expansion of terms. The discussion remains focused on the expression without resolving the mathematical steps or definitions involved.
RTCNTC said: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?
RTCNTC said: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.
kaliprasad said: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 $(a-b)(a+b)$
RTCNTC said:In that case, can you work it out for me step by step?
kaliprasad said:your steps are correct till you instead of expanding $(a-b)^2$ did $(a-b)(a+b) $ and I work it out here
$27-(a-b)^3$
= $3^3-(a-b)^3$
$= (3-(a-b))(3 + (a-b) + (a-b)^2)$
MarkFL said:This last step should be:
$$\left(3-(a-b)\right)\left(3^2+3(a-b)+(a-b)^2\right)$$ :D
MarkFL said:This last step should be:
$$\left(3-(a-b)\right)\left(3^2+3(a-b)+(a-b)^2\right)$$ :D