Solve Factorization of ##a^6-b^6## - Different Approach

[chwala]

Homework Statement

factorize $a^6-b^6$

Relevant Equations

factorization

$a^6-b^6$≡$(a^3-b^3)(a^3+b^3)$
≡$(a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2)$
and
$(a^2-ab+b^2)(a^2+ab+b^2)≡(a^4+2a^2b^2+b^4)$
letting $a^2=x, b^2=y$
$a^4+2a^2b^2+b^4= x^2+2xy+y^2)$
=$(x+y)(x+y)$
thus,
$a^6-b^6$=$(a-b)(a+b)(a^2+b^2)(a^2+b^2)$

is there a different approach to this?

 

[Vanadium 50]

You might want to plug in some values for a and b to check.

 

[LCKurtz]

Homework Statement:: factorize $a^6-b^6$
Relevant Equations:: factorization

$a^6-b^6$≡$(a^3-b^3)(a^3+b^3)$
≡$(a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2)$

I would stop right there. It's factored.

 

[chwala]

I would stop right there. It's factored.

Thanks, yeah right...i know i am right i thought the math wizards here might have a different approach to it...Thanks LCkurtz.

 

[Mark44]

Homework Statement:: factorize $a^6-b^6$
Relevant Equations:: factorization

$a^6-b^6$≡$(a^3-b^3)(a^3+b^3)$
≡$(a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2)$

@LCKurtz is correct. Stop here -- it's factored.

and
$(a^2-ab+b^2)(a^2+ab+b^2)≡(a^4+2a^2b^2+b^4)$

No, the middle term on the right side is incorrect. You should have gotten $a^4 + a^2b^2 + b^4$

letting $a^2=x, b^2=y$
$a^4+2a^2b^2+b^4= x^2+2xy+y^2)$
=$(x+y)(x+y)$
thus,
$a^6-b^6$=$(a-b)(a+b)(a^2+b^2)(a^2+b^2)$

is there a different approach to this?

 

[Klystron]

Homework Statement:: factorize $a^6-b^6$
{snip}
is there a different approach to this?

Perhaps you are thinking of the binomial theorem?

https://en.wikipedia.org/wiki/Binomial_theorem

 

[chwala]

yes, my expansion is wrong. I did this in a hurry oops...

 

[chwala]

Perhaps you are thinking of the binomial theorem?

interesting, i do not see how one can use binomal expansion on this, kindly show me how...i had thought of that initially...

 

[chwala]

Perhaps you are thinking of the binomial theorem?

i am still waiting for your approach ...

 

[Mark44]

Perhaps you are thinking of the binomial theorem?

i am still waiting for your approach ...

Unless I'm missing something, I don't see how the binomial theorem is applicable here. It deals with a sum or difference of variables raised to some power, not the sum or difference of powers. IOW, the domain of application for the binomial theorem is expressions such as $(a \pm b)^n$.

As already noted, your first factorization in post #1 is correct, and the second one is incorrect. Once you have found the prime factors of a number or an algebraic expression, that factorization is unique, so why should there necessarily be another approach?

 

[Klystron]

Since you had already factored the equation I took your secondary query as request for additional comments.

I recommend following the approach in post #2 to insert values in a and b with the implied idea to plot the results to improve understanding. You could write a small generic program to solve and plot similar equations for future use and also to test your expanded equation.

I agree with you that equations of this form resemble polynomials that fall within binomial theory but the encyclopedia reference was also intended as reminder on technique including expansion, geometric analysis, generalizations, and application of transcendental functions particularly natural logarithms. Thanks.

[EDIT 20200218: removed irrelevant comments.]

 

[vanhees71]

Just calculate the roots. Any polynomial factorizes into linear factors (over $\mathbb{C}$ of course, not $\mathbb{R}$ though). Here it's very easy
$$a^6-b^6=0$$
has the independent solutions
$$a_k=b \exp(2 \pi \mathrm{i} k/6)=b \exp(\pi \mathrm{i} k/3), \quad k \in \{0,1,2,3,4,5 \}.$$
The factorization reads
$$a^6-b^6=\prod_{k=0}^5 (a-a_k).$$