Question

Factorise (4x-5y) cube + (5y-6z) cube + ( 6z-4x) cube

Answer 1

Step-by-step explanation:

(4x-5y)³+(5y-6z)³+(6z-4x)³

let (4x-5y) be a, (5y-6z) be b and (6z-4x) be c.

If a+b+c=0, then a³+b³+c³=3abc.

4x-5y+5y-6z+6z-4x=0

Thus, (4x-5y)³+(5y-6z)³+(6z-4x)³= 3(4x-5y)(5y-6z)(6z-4x)

=(4x-5y)³+(5y-6z)³+(6z-4x)³={(4x-5y)(5y-6z)}(6z-4x)

=(4x-5y)³+(5y-6z)³+(6z-4x)³={20xy-24xz-25y²+30yz}(6z-4x)

=(4x-5y)³+(5y-6z)³+(6z-4x)³=120xyz-120xyz-144xz²+96x²z-150y²z+180yz²-80x²y+100xy²

=(4x-5y)³+(5y-6z)³+(6z-4x)³=48xz(-3z+2x)+30yz(-5y+6z)+20xy(-4x+5y)