Question : If $x=222, y=223$ and $z=224$, then find the value of $x^3+y^3+z^3-3 x y z$.
Option 1: 2007
Option 2: 2004
Option 3: 2006
Option 4: 2005
Answer (1)
16th Jan, 2024
Correct Answer: 2007
Solution :
$x^3 + y^3 +z^3-3xyz=\frac{1}{2}(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]$
$= (\frac{1}{2})(222+ 223 + 224)[(222-223)^2 + (223-224)^2 + (224-222)^2]$
$= (\frac{1}{2})(669)[(1)+(1) + (4)]$
$= 3\times 669 = 2007$
Hence, the correct answer is 2007.
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