If $\mathrm{ x}=110,y=111,z=112,$  then find the value of ${\mathrm{ x}}^3+y^3+z^3-3xyz.$

The correct answer is 999.

Given the values:

$x=110$

$y=111$

$z=112$

We need to find the value of the algebraic expression:

$x^3+y^3+z^3-3xyz$

We can use the standard algebraic identity:

$x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)[(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]$

Step 1: Compute the sum
$(x+y+z)$
:

$x+y+z=110+111+112=333$

Step 2: Compute the differences between consecutive terms and square them:

$x-y=110-111=-1\Rightarrow (x-y{)}^2=(-1{)}^2=1$

$y-z=111-112=-1\Rightarrow (y-z{)}^2=(-1{)}^2=1$

$z-x=112-110=2\Rightarrow (z-x{)}^2=2^2=4$

Step 3: Add the squared differences together:

$(x-y{)}^2+(y-z{)}^2+(z-x{)}^2=1+1+4=6$

Step 4: Substitute the results back into the identity:

$x^3+y^3+z^3-3xyz=\frac{1}{2}\times 333\times 6$

$x^3+y^3+z^3-3xyz=333\times 3=999$

Thus, the final value of the given expression is 999.