Factorise: (x-2y)^3 + (2y-3x)^3 + (3x-x)^3
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Step-by-step explanation:
Correct option is
B
3(x−2y)(2y−3z)(3z−x)
We know that,
a
3
+b
3
+c
3
=(a+b+c)(a
2
+b
2
+c
2
−ab−bc−ca)+3abc
If a+b+c=0, then a
3
+b
3
+c
3
=3abc
Now, given (x−2y)
3
+(2y−3z)
3
+(3z−x)
3
Here, a=(x−2y), b=(2y−3z), c=(3z−x) and a+b+c=(x−2y)+(2y−3z)+(3z−x)=0
Therefore
(x−2y)
3
+(2y−3z)
3
+(3z−x)
3
=3(x−2y)(2y−3z)(3z−x)
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