If $x+y+z=2, x^3+y^3+z^3-3 x y z=74$, then $\left(x^2+y^2+z^2\right)$ is equal to :

26

We know that,

a³ + b³ + c³ - 3abc = \(\frac{(a + b + c)}{2}\) [3(a² + b² + c²) - (a + b + c)²]

x + y + z = 2

x³ + y³ + z³ - 3xyz = 74

According to the question,

\(\frac{2}{2}\) [3(x² + y² + z²) - 2²] = 74

= 3(x² + y² + z²) = 74 + 4 = 78

= x² + y² + z²= 26