How to Factor Cubic Terms in Algebraic Expressions?

[mathdad]

Factor the expression.

(a - a^2)^3 + (a^2 - 1)^3 + (1 - a)^3

(a - a^2)(a - a^2)(a - a^2) + (a^2 - 1)(a^2 - 1)(a^2 - 1) +
(1 - a)(1 - a)(1 -a)

Is this the right approach thus far?

 

[MarkFL]

I would first look at factoring each expression:

\(\displaystyle (a-a^2)^3+(a^2-1)^3+(1-a)^3=a^3(1-a)^3-(a+1)^3(1-a)^3+(1-a)^3\)

Now, we have a factor common to all 3 terms...;)

 

[mathdad]

Where did a^3 come from?

Factor out (1 - a)^3.

(1 - a)^3[a^3 - (1 - a)^3 + 1]

Inside the brackets, I must apply the difference of cubes
to a^3 - (1 - a)^3, right?

 

[skeeter]

Where did a^3 come from?

first term ...

$(a - a^2)^3 + (a^2 - 1)^3 + (1 - a)^3$

$[a(1-a)]^3 + [(a-1)(a+1)]^3 + (1-a)^3$

$a^3(1-a)^3 - (a+1)^3(1-a)^3 + (1-a)^3$

continuing ...

$(1-a)^3[(a^3 + 1) - (a+1)^3]$

$(1-a)^3[(a+1)(a^2-a+1) - (a+1)^3]$

$(1-a)^3(a+1)[(a^2-a+1) - (a+1)^2]$

$(1-a)^3(a+1)(-3a)$

$3a(a-1)^3(a+1)$

 

[mathdad]

I see that this is not an average factoring problem.