1. **Stating the problem:** We are given a system of three equations in four variables $(x,y,z,t)$: $$\begin{cases} x^2 + y^3 + z^3 + t^3 = 0 \\ x^2 + y^2 + z^2 + t = 2 \\ x + y + z + t = 0 \end{cases}$$ and a point $(0,-1,1,0)$ that satisfies these equations. We want to understand the implicit function $\psi : t \to (x(t), y(t), z(t), t)$ defined by these equations near this point. 2. **Check the point satisfies the system:** - For $(0,-1,1,0)$: - $x^2 + y^3 + z^3 + t^3 = 0^2 + (-1)^3 + 1^3 + 0^3 = -1 + 1 = 0$ - $x^2 + y^2 + z^2 + t = 0^2 + 1 + 1 + 0 = 2$ - $x + y + z + t = 0 -1 + 1 + 0 = 0$ So the point satisfies the system. 3. **Use the Implicit Function Theorem:** We want to solve for $(x,y,z)$ as functions of $t$ near $t=0$. Define the function $F: \mathbb{R}^4 \to \mathbb{R}^3$ by $$F(x,y,z,t) = \begin{pmatrix} x^2 + y^3 + z^3 + t^3 \\ x^2 + y^2 + z^2 + t - 2 \\ x + y + z + t \end{pmatrix}.$$ 4. **Compute the Jacobian matrix of $F$ with respect to $(x,y,z)$ at the point $(0,-1,1,0)$:** $$J = \begin{pmatrix} \frac{\partial F_1}{\partial x} & \frac{\partial F_1}{\partial y} & \frac{\partial F_1}{\partial z} \\ \frac{\partial F_2}{\partial x} & \frac{\partial F_2}{\partial y} & \frac{\partial F_2}{\partial z} \\ \frac{\partial F_3}{\partial x} & \frac{\partial F_3}{\partial y} & \frac{\partial F_3}{\partial z} \end{pmatrix}.$$ Calculate each partial derivative: - $\frac{\partial F_1}{\partial x} = 2x$, at $(0,-1,1,0)$ is $0$ - $\frac{\partial F_1}{\partial y} = 3y^2$, at $y=-1$ is $3$ - $\frac{\partial F_1}{\partial z} = 3z^2$, at $z=1$ is $3$ - $\frac{\partial F_2}{\partial x} = 2x$, at $0$ is $0$ - $\frac{\partial F_2}{\partial y} = 2y$, at $-1$ is $-2$ - $\frac{\partial F_2}{\partial z} = 2z$, at $1$ is $2$ - $\frac{\partial F_3}{\partial x} = 1$ - $\frac{\partial F_3}{\partial y} = 1$ - $\frac{\partial F_3}{\partial z} = 1$ So $$J = \begin{pmatrix} 0 & 3 & 3 \\ 0 & -2 & 2 \\ 1 & 1 & 1 \end{pmatrix}.$$ 5. **Check if $J$ is invertible:** Calculate $\det(J)$: $$\det(J) = 0 \cdot (-2 \cdot 1 - 2 \cdot 1) - 3 \cdot (0 \cdot 1 - 2 \cdot 1) + 3 \cdot (0 \cdot 1 - (-2) \cdot 1) = 0 - 3 \cdot (-2) + 3 \cdot 2 = 6 + 6 = 12 \neq 0.$$ Since $\det(J) \neq 0$, by the Implicit Function Theorem, there exist unique differentiable functions $x(t), y(t), z(t)$ near $t=0$ such that $F(x(t), y(t), z(t), t) = 0$. 6. **Summary:** The system implicitly defines $x,y,z$ as functions of $t$ near the point $(0,-1,1,0)$. **Final answer:** The Jacobian matrix with respect to $(x,y,z)$ at $(0,-1,1,0)$ is invertible, so the implicit function $\psi : t \to (x(t), y(t), z(t), t)$ exists and is differentiable near $t=0$.