Question

factorise (2x-3y)^3+(3y-4z)^3+(4z-2x)^3

Answer 1

Answer:

Step-by-step explanation:

(2x-3y)3+(3y-4z)3+2^3(2z-x)3

=(2x-3y)^3+(3y-4z)^3+(4z-2x)^3

in last term we taken 2 into inside

[if a+b+c=0 then a3+b3+c3=3abc]

here a=2x-3y

b=3y-4z

c=4z-2x

a+b+c =2x-3y+3y-2x+4z-2x

=0

therefore

(2x-3y)3+(3y-4z)3+(4z-2x)3

= 3*(2x-3y)(3y-4z)(4z-2x)

=3*2(2x-3y)(3y-4z)(2z-x)

Answer 2

Answer:

Step-by-step explanation:

(2x-3y)3+(3y-4z)3+2^3(2z-x)3

=(2x-3y)^3+(3y-4z)^3+(4z-2x)^3

in the last term, we took 2 into the inside

[if a+b+c=0 then a3+b3+c3=3abc]

here a=2x-3y

b=3y-4z

c=4z-2x

a+b+c =2x-3y+3y-2x+4z-2x

=0

therefore

(2x-3y)3+(3y-4z)3+(4z-2x)3

= 3*(2x-3y)(3y-4z)(4z-2x)

=3*2(2x-3y)(3y-4z)(2z-x)