Thread: Factorising an equation

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

Originally Posted by Shaun Gill

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

What does a.c (or a.b, b.a ...) means?

sorry a.b means a multiplied by b

Originally Posted by Shaun Gill

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)

Originally Posted by Soroban

[size=3]Hello, Shaun!

Is there a typo?
If the fourth term is positive, I can factor it

I thought also that there is typo.

Only if...