can someone show me how to go about factorising this please?
(a.c^3)-(a.b^3)+(b.a^3)-(b.c^3)-(c.a^3)-(c.b^3)
cheers
There are probably a million different ways to go about factoring pieces of it. I'm not sure there you can find any sort of complete factorization of it (a factorization that breaks it down into multiplicative factors only).
The best I've been able to do so far is:
a^3b - a^3c - ab^3 + ac^3 - b^3c - bc^3
a^3(b - c) - a(b^3 - c^3) - bc(b^2 + c^2)
a^3(b - c) - a(b - c)(b^2 + c^2 + bc) - bc(b^2 + c^2)
a(b - c)(a^2 - b^2 - c^2 - bc) - bc(b^2 + c^2)
-Dan
Hello, Shaun!
Is there a typo?
If the fourth term is positive, I can factor it.
ac³ - ab³ +a³b + bc³ - a³c - b³c
We have: .a³b - a³c - ab³ + ac³ - b³c + bc³
Then: .a³(b - c) - a(b³ - c³) - bc(b² - c²)
. . = .a³(b - c) - a(b - c)(b² + bc + c²) - bc(b - c)(b + c)
Factor out (b - c):
. . (b - c) [a³ - a(b² + bc + c²) - bc(b + c)]
. . = .(b - c) [a³ - ab² - abc - ac² - b²c -bc²]
. . = .(b - c)[a³ - ac² - abc - bc² - ab² - b²c]
. . = .(b - c) [a(a² - c²) - bc(a - c) - b²(a - c)]
. . = .(b - c) [a(a - c)(a + c) - bc(a - c) - b²(a - c)]
Factor out (a - c):
. . (b - c)(a - c) [a(a + c) - bc - b²]
. . = .(b - c)(a - c) [a² + ac - bc - b²]
. . = .(b - c)(a - c) [a² - b² + ac - bc]
. . = .(b - c)(a - c) [(a - b)(a + b) + c(a - b)]
Factor out (a - b):
. . (a - b)(b - c)(a - c)(a + b + c)