If x, y, z are real numbers such that $x+y+z=10$ and $xy+yz+zx=18$, then what is the value of $x{}^3+y{}^3+z{}^3-3xyz$?

Given:

x + y + z = 10

xy + yz + zx = 18

Formula used:

The expression for the sum of cubes of three numbers can be expressed as:

x³ + y³ + z³ - 3xyz = (x + y + z) [(x + y + z)² - 3(xy + yz + zx)]

Calculation:

Substituting the given values:

x³ + y³ + z³ - 3xyz = (10) [(10)² - 3(18)]

⇒ x³ + y³ + z³ - 3xyz = 10 [100 - 54]

⇒ x³ + y³ + z³ - 3xyz = 10 × 46

⇒ x³ + y³ + z³ - 3xyz = 460

∴ The correct answer is option (3).