If x = 222, y = 223 and z = 224, then find the value of x³ + y³ + z³ – 3xyz
Given- x = 222, y = 223, z = 224 Then, as we know that – x ³ + y³ +z³ –3xyz = (x +y + z)/2 [(x-y) ² +(y-z) ² +(z-x) ²] = (222+223+224)/2 [1+1+4] = (669) ×3 = 2007 x ³ + y³ +z³ –3xyz =2007
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