The first equation implies $x^3 = -y^2 - 1$. Thus $x^3 < 0$, implying also $x < 0$. Similarly from the second, third and fourth equations we obtain $y < 0$, $z < 0$ and $t < 0$, respectively. The sum of negative numbers $x$, $y$, $z$, $t$ is negative, contradicting the fifth equation.

Suppose that the system has a solution. W.l.o.g., let $x$ be variable with the largest value. Then $4x \ge x + y + z + t$, which by the last equation implies $x \ge 0$. Consequently, also $x^3 \ge 0$. As $y^2 \ge 0$, this implies $1 + x^3 + y^2 \ge 1$, contradicting the first equation. Hence no solution can exist.