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

Correct option is A

Given:

x = 110, y = 111, z = 112

We need to find the value of:

​$$x^3 + y^3 + z^3 - 3xyz$$​

Formula Used:

​$$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$$​​

Solution:

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

​$$x^2 + y^2 + z^2 = 110^2 + 111^2 + 112^2 = 12100 + 12321 + 12544 = 36965$$​​

xy + yz + zx = 12210 + 12432 + 12320 = 36962

​$$x^2 + y^2 + z^2 - (xy + yz + zx) = 36965 - 36962 = 3$$​​

​$$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) = 333 \times 3 = 999$$​​

Thus, the value of $$x^3 + y^3 + z^3 - 3xyz$$​ is 999. 

Alternate Solution: ​

Using formula;

​$$x^3 + y^3 + z^3 - 3xyz = \frac12(x + y + z)[(x-y)^2 + (y-z)^2 + (z-x)^2]$$ 

​$$= \frac12(110 + 111 + 112)[(110-111)^2 + (111-112)^2 + (112-110)^2] \\ \ \\ =\frac12(333)[(1)^2+(1)^2+(2)^2]\\ \ \\ = \frac12(333)[6]\\ \ \\ =333\times 3\\ \ \\ = 999$$​​