factorise: (x-y)^2 + (y-z)^3 + (z-x)^3
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We know that
a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ac)
when a + b + c = 0
a³ + b³ + c³ - 3abc = 0
=> a³ + b³ + c³ = 3abc
Let a = x - y
b = y - z
c = z - x
a + b + c = x - y + y - z + z - x
=> the above equation turns out to be 3(x - y)(y - z)(z - x)
a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ac)
when a + b + c = 0
a³ + b³ + c³ - 3abc = 0
=> a³ + b³ + c³ = 3abc
Let a = x - y
b = y - z
c = z - x
a + b + c = x - y + y - z + z - x
=> the above equation turns out to be 3(x - y)(y - z)(z - x)
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