solve x3+y3+z3-3xyz? when x=2,y=1,z=(-3)
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x^3 + y^3 + z^3 -3xyz [ x = 2 , y = 1 , z = -3 ]
substitute the x , y ,z value in the equation
= (2)^3 + ( 1 )^3 + (-3)^3 - 3×2×1×-3
= 8 + 1 - 27 + 18
= 8 + 1 + 18 - 27
= 9 + 18 - 27
= 27 - 27
= 0
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