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
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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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Question : $\text { If } x^2+y^2+z^2=x y+y z+z x \text { and } x=1 \text {, then find the value of } \frac{10 x^4+5 y^4+7 z^4}{13 x^2 y^2+6 y^2 z^2+3 z^2 x^2}$.
Option 1: 2
Option 2: 0
Option 3: –1
Option 4: 1
Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{(y z)}+\frac{y^2}{(x z)}+\frac{z^2}{(x y)}$?
Option 1: 1
Option 3: 2
Option 4: 3
Question : What is the simplified value of $\frac{(x+y+z)(x y+y z+z x)–x y z}{(x+y)(y+z)(z+x)}$?
Option 1: $y$
Option 2: $z$
Option 3: $1$
Option 4: $x$
Question : If $x+y+z=0$ and $x^2+y^2+z^2=40$, then what is the value of $x y+y z+z x?$
Option 1: –20
Option 2: 5
Option 3: –5
Option 4: –10
Question : If $\frac{1}{x+\frac{1}{y+\frac{2}{z+\frac{1}{4}}}}=\frac{29}{79}$, where x, y, and z are natural numbers, then the value of $(2 x+3 y-z)$ is:
Option 1: 0
Option 2: 4
Option 3: 1
Option 4: 2
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