What are the factors of (2x-3y )3 + (3y-4z) 3 + 8(2z-x) 3
Answers
Answered by
29
(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)
=(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)
Answered by
9
(2x-3y)3 + (3y-4z) 3 + 8 (2z-x) 3
= 3 (2x-3y+3y-4z+16z-8x)
= 3 (-6x+12z)
= 3 . 6 (2z-x)
= 18 (2z-x)
= 3 (2x-3y+3y-4z+16z-8x)
= 3 (-6x+12z)
= 3 . 6 (2z-x)
= 18 (2z-x)
mysticd:
check ur solution i think ur mistaken
In the question, representation of powers is in confusion. Thanks.
ur right , that is main problem in the typing and we get
confused
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