Let x and y = the numbers

$x + y = 2$ → Equation (1)

$1 + y' = 0$

$y' = -1$

$z = x^3 + y^2$ → Equation (2)

$dz/dx = 3x^2 + 2y \, y' = 0$

$3x^2 + 2y(-1) = 0$

$y = \frac{3}{2}x^2$

From Equation (1)

$x + \frac{3}{2}x^2 = 2$

$2x + 3x^2 = 4$

$3x^2 + 2x - 4 = 0$

$x = 0.8685 \, \text{&} \, -1.5352$

Use

$x = 0.8685$

$y = \frac{3}{2}(0.8685^2)$

$y = 1.1315$

$z = 0.8685^3 + 1.1315^2$

$z = 1.9354$ *answer*