Question
If x = y = 333 and z = 334, then the value of x3 + y3 + z3 - 3xyz is?
Answer: Option C
x = y = 333, z = 334
⇒ x3 + y3 + z3 - 3xyz = $$\frac{1}{2}$$ (x + y + z) [(x - y)2 + (y - z)2 + (z - x)2]
⇒ x3 + y3 + z3 - 3xyz = $$\frac{1}{2}$$ (333 + 333 + 334) (333 - 333)2 + (333 - 334)2 + (334 - 333)2
⇒ x3 + y3 + z3 - 3xyz = $$\frac{1}{2}$$ (1000) (0 + 1 + 1)
⇒ x3 + y3 + z3 - 3xyz = 1000
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x = y = 333, z = 334
⇒ x3 + y3 + z3 - 3xyz = $$\frac{1}{2}$$ (x + y + z) [(x - y)2 + (y - z)2 + (z - x)2]
⇒ x3 + y3 + z3 - 3xyz = $$\frac{1}{2}$$ (333 + 333 + 334) (333 - 333)2 + (333 - 334)2 + (334 - 333)2
⇒ x3 + y3 + z3 - 3xyz = $$\frac{1}{2}$$ (1000) (0 + 1 + 1)
⇒ x3 + y3 + z3 - 3xyz = 1000
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