MHB How to Factor Cubic Terms in Algebraic Expressions?

AI Thread Summary
The discussion focuses on factoring the expression (a - a^2)^3 + (a^2 - 1)^3 + (1 - a)^3. The initial approach involves recognizing common factors and applying the difference of cubes. The expression simplifies to (1 - a)^3[a^3 - (1 - a)^3 + 1], leading to further factoring. Ultimately, the final result is expressed as 3a(a - 1)^3(a + 1). This demonstrates that the problem requires more than basic factoring techniques.
mathdad
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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?
 
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I would first look at factoring each expression:

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

Now, we have a factor common to all 3 terms...;)
 
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?
 
RTCNTC said:
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 ...

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

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

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

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

$3a(a-1)^3(a+1)$
 
I see that this is not an average factoring problem.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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