Factorise:(x-y)^3+(y-z)^3+(z-x)^3
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2
Step by step explanation:-
When a+b+c = 0 we know that a³ + b³ + c³ = 3 a b c
(x-y)³ + (y-z)³ + (z - x)³
= 3 (x - y) (y -z) (z- x)
= 3 (xy - x z -y² + yz) (z -x)
= 3 (xyz - x² y - x z² + x² z - y² z + x y² + y z² - xyz)
= 3 [ x² (z - y) + z² (y - x) + y² (x - z) ]
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similarly x - 2y + 2y - 3z + 3z - x = 0
hence, (x-2y)³ + (2y - 3z)³ + (3z-x)³ = 3 (x -2y) (2y -3z)(3z - x)
Answered by
12
a+b+c = 0
a³ + b³ + c³ = 3 a b c
(x-y)³ + (y-z)³ + (z - x)³
= 3 (x - y) (y -z) (z- x)
= 3 (xy - x z -y² + yz) (z -x)
= 3 (xyz - x² y - x z² + x² z - y² z + x y² + y z² - xyz)
= 3 [ x² (z - y) + z² (y - x) + y² (x - z) ]
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similarly x - 2y + 2y - 3z + 3z - x = 0
hence, (x-2y)³ + (2y - 3z)³ + (3z-x)³ = 3 (x -2y) (2y -3z)(3z - x)
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