$1.54 \sqrt[3]{38+\sqrt{1445}}+\sqrt[3]{38-\sqrt{1445}}=4$.
### 1.54 Let

$\sqrt[3]{38+\sqrt{1445}}+\sqrt[3]{38-\sqrt{1445}}=x$.

Cubing both sides of this equation, and using formula (1.11), we get

$38+\sqrt{1445}+38-\sqrt{1445}+3 \sqrt[3]{(38+\sqrt{1445})(38-\sqrt{1445})} x=x^{3}$,

or $x^{3}+3 x-76=0$. By substitution, we verify that $x=4$ is one of the roots of the resulting cubic equation: $64+12-76=0$.

Transform this cubic equation:

$$
\begin{aligned}
& x^{3}-64=3(4-x) ;(x-4)\left(x^{2}+4 x+16\right)+3(x-4)=0 \\
& (x-4)\left(x^{2}+4 x+19\right)=0
\end{aligned}
$$

But the factor $x^{2}+4 x+19$ has no real roots. Therefore, 4 is the only possible real value for $x$, which proves the required equality (since it is obvious that $\sqrt[3]{38+\sqrt{1445}}+\sqrt[3]{38-\sqrt{1445}}$ is a real number).