Question : If $x=y=333$ and $z=334$, the value of $x^3+y^3+z^3-3xyz$ is:
Option 1: $0$
Option 2: $667$
Option 3: $1000$
Option 4: $2334$
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Correct Answer: $1000$
Solution :
Given: $x=y=333$ and $z=334$
$x^3+y^3+z^3-3xyz$
= $\frac{1}{2}(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]$
= $\frac{1}{2}×(333+333+334)(0+1+1)$
= $1000$
Hence, the correct answer is $1000$.
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