factorize it plzzzzz
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Hii..
Here is your answer...
Solution:-
Since we know that,
a^3+b^3+c^3-3abc=(a+b+C)(a^2+b^2+c^2-ab-bc-ca)
If (a+B+C)=0
So,
a^3+b^3+c^3=3abc
Now here,
a=x-2y
B=2y-3z
c=3z-x
So,
(x-2y)(2y-3z)(3z-x)=0
Therefore,
(x-2y)^3(2y-3z)^3(3z-x)^3=3(x-2y)(2y-3z)(3z-x)
Hope it helps uh...✌️✌️✌️
Here is your answer...
Solution:-
Since we know that,
a^3+b^3+c^3-3abc=(a+b+C)(a^2+b^2+c^2-ab-bc-ca)
If (a+B+C)=0
So,
a^3+b^3+c^3=3abc
Now here,
a=x-2y
B=2y-3z
c=3z-x
So,
(x-2y)(2y-3z)(3z-x)=0
Therefore,
(x-2y)^3(2y-3z)^3(3z-x)^3=3(x-2y)(2y-3z)(3z-x)
Hope it helps uh...✌️✌️✌️
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