[Algebra]
Problem: Let $T=0$. Compute the real value of $x$ for which there exists a solution to the system of equations

$$
\begin{aligned}
x+y & =0 \\
x^{3}-y^{3} & =54+T .
\end{aligned}
$$

Solution: $\quad$ Plug $y=-x$ into the second equation to obtain $x=\sqrt[3]{\frac{54+T}{2}}$. With $T=0, x=\sqrt[3]{27}=3$.

Final answer: 3