Please explain and factorize this:b^15 - b^10c^2 - 1/81b^5c^4+1/81c^6?

Answer 1

This is factored by grouping.
b^15-b^10c^2 -1/81b^5c^4+1/81c^6
b^10(b^5-c^2)-1/81c^4(b^5-c^2)
(b^10-1/81c^4)(b^5-c^2)
The first bracketed term is the difference of 2 squares
(b^5-1/9c^2)(b^5+1/9c^2)(b^5-c^2)
A messy little thing isn't it!

Answer 2

b^15 - b^10c^2 - 1/81b^5c^4+1/81c^6

Play with this one a little bit. I started by factoring b^10 out of the first 2 terms, and -1/81 c^4 out of the secont two terms:
b^10 (b^5 - c^2) - 1/81 c^4 ( b^5 - c^2)
Now, we can factor b^5 - c^2 out of each term:
( b^5 - c^2) (b^10 - 1/81 c^4)
We're done with the first factor, but the second factor (b^10 - 1/81 c^4) is the difference of 2 perfect squares, so we can write:
( b^5 - c^2) (b^5 + 1/9 c^2) (b^5 - 1/9 c^2)
Now, I believe we have done more than may be necessary in practical terms, but this may satisfy the writer of the question.
Sometimes this stuff can just takes some lead, paper, time, and effort to solve. I didn't have a clue about where to start before I wrote the expression on paper; then I just started doing something with it. Paper is cheap, and I figure I can always neaten it up or correct something if I have something to work with. Have fun with it.

Answer 3

Let x = b^5
and y = c^2
then the expression becomes
x^3 - x^2 y - 1/81 x y^2 + 1/81 y^3
which factor to
x^2 (x - y) - 1/81 y^2 (x - y)
= (x - y) (x^2 - 1/81 y^2)
= (x - y) (x - y/9) (x + y/9)
= (b^5 - c^2) ( b^5 - (1/9)c^2 ) ( b^5 + (1/9)c^2 )

How's that!

Answer 4

[02]
b^15-b^10c^2-1/81 b^5c^4 +1/81 c^6
=b^10(b^5-c^2)-1/81 c^4(b^5-c^2)
=(b^5-c^2)(b^10-1/81 c^4)
=(b^5-c^2){(b^5)^2-(1/9 c^2)^2}
=(b^5-c^2)(b^5+1/9 c^2)(b^5 -1/9 c^2)

Answer 5

Trial and error.

Answer 6

b^15-b^10c^2-1/81b^5c^4+1/81c^6=
=b^10(b^5-c^2)-1/81c^4(b^5-c^2)=
=(b^5-c^2)(b^10-1/81c^4)=(b^5-c^2)((b^5)^2-(1/9c^2)^2)=
=(b^5-c^2)(b^5-1/9c^2)(b^5+1/9c^2)