plz factorise this equation without finding cubes
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Just multiply in the last step...Hope it helped u
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(x-2y)³+(2y-3z)³+(3z-x)³
a=x-2y
b=2y-3z
c=3z-x
If a+b+c= 0, then a³+b³+c³=3abc
a+b+c= x-2y+2y-3z+3z-x
=x-x+2y-2y+3z-3z
=0
a³+b³+c³=3abc
=3(x-2y)(2y-3z)(3z-x)
Hope it helps ☺
Please mark as Brainliest
a=x-2y
b=2y-3z
c=3z-x
If a+b+c= 0, then a³+b³+c³=3abc
a+b+c= x-2y+2y-3z+3z-x
=x-x+2y-2y+3z-3z
=0
a³+b³+c³=3abc
=3(x-2y)(2y-3z)(3z-x)
Hope it helps ☺
Please mark as Brainliest
abhinav101:
do is dis d do final factorisation??
R u trying to find the values of x y nd z ???
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