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 :

24 26 29 22

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