# Absurdly Complicated (to me) Factoring Problem

• Aug 20th 2009, 01:39 PM
leveebreaker05
Absurdly Complicated (to me) Factoring Problem
I was given a review handout for a Calc. + Anal. Geom. (lol!) and I ran into an inconspicuously confusing factoring problem:

54ał - 16bł

I factored it to 2(27ał - 8bł), but he got to
2(3a - 2b)(9a˛ + 6ab + 4b˛)

May someone please explain to me how solving this this was expected to be innate algebraic practicality?

…or please explain how he might have gotten to his solution. From there, I hope I'd be able to tackle almost any similarly ridonque. factoring problem…
• Aug 20th 2009, 01:45 PM
skeeter
Quote:

Originally Posted by leveebreaker05
I was given a review handout for a Calc. + Anal. Geom. (lol!) and I ran into an inconspicuously confusing factoring problem:

54ał - 16bł

I factored it to 2(27ał - 8bł), but he got to
2(3a - 2b)(9a˛ + 6ab + 4b˛)

May someone please explain to me how solving this this was expected to be innate algebraic practicality?

…or please explain how he might have gotten to his solution. From there, I hope I'd be able to tackle almost any similarly ridonque. factoring problem…

pattern for factoring the difference of cubes ...

$\displaystyle x^3 - y^3 = (x-y)(x^2 + xy + y^2)$

in your problem, $\displaystyle x = 3a$ and $\displaystyle y = 2b$
• Aug 20th 2009, 01:58 PM
HallsofIvy
More generally, it is true that
$\displaystyle x^n- y^n= (x- y)(x^{n-1}+ x^{n-2}y+ x^{n-3}y^2+ \cdot\cdot\cdot\+ x^2y^{n- 3}+ xy^{n-2}+ y^{n-1})$
and, if n is odd, that
$\displaystyle x^n+ y^n= (x+ y)(x^{n-1}- x^{n-2}y+ x^{n-3}y^2+ \cdot\cdot\cdot\- x^2y^{n- 3}+ xy^{n-2}- y^{n-1})$
• Aug 20th 2009, 03:30 PM
stapel
Quote:

Originally Posted by leveebreaker05
May someone please explain to me how...he might have gotten to his solution.

He applied the formula for factoring a difference of cubes. This is a formula that you should memorize! (Wink)
• Aug 21st 2009, 09:27 AM
samtcs
In my opinion,

54(a^3)-16(b^3) = 2[27(a^3)-8(b^3)]

Then, we notice that 27^(1/3)=3, and -8^(1/3)=-2.
Then,
(3a-2b)^3 = 27(a^3)-8(b^3)-54(a^2)(b)+36(a)(b^2)
27(a^3)-8(b^3)=(3a-2b)^3+54(a^2)(b)-36(a)(b^2)

Then,
54(a^3)-16(b^3) = 2[(3a-2b)^3+54(a^2)(b)-36(a)(b^2)]
= 2[(3a-2b)^3+18ab(3a-2b)]
= 2[(3a-2b)((3a-2b)^2+18ab)]
= 2(3a-2b)(9(a^2)+6ab+4(b^2))
That's the answer....but i think maybe got some better solution...
• Aug 21st 2009, 01:17 PM
Wilmer