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…
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.
(3a-2b)^3 = 27(a^3)-8(b^3)-54(a^2)(b)+36(a)(b^2)
54(a^3)-16(b^3) = 2[(3a-2b)^3+54(a^2)(b)-36(a)(b^2)]
That's the answer....but i think maybe got some better solution...