Initial 34. Solve the equation in the set of real numbers
$$
\sqrt[3]{(x+14)(x-16)}+\sqrt[5]{(x+15)(17-x)}=1 .
$$
Let $a=\sqrt[5]{(x+14)(x-16)}$,
$$
b=\sqrt[5]{(x+15)(17-x)} \text {. }
$$

Then we have
$$
\left\{\begin{array}{l}
a+b=1 \\
a^{5}+b^{5}=31
\end{array}\right.
$$
(2) $\div$ (1) gives $a^{4}+b^{4}-a^{3} b-a b^{3}+a^{2} b^{2}=31$.
From (1), squaring and rearranging gives $a^{2}+b^{2}=1 - 2ab$.
From (4), squaring and rearranging gives $a^{4}+b^{4}=2a^{2}b^{2}-4ab+1$.
Substituting (4) and (5) into (3) yields
$$
2a^{2}b^{2}-4ab+1-ab(1-2ab)+a^{2}b^{2}=31 .
$$

Simplifying, we get $a^{2}b^{2}-ab-6=0$.
Treating $ab$ as an unknown, we get $ab=-2, ab=3$. Thus,
$$
\left\{\begin{array}{l}
a+b=1, \\
ab=-2,
\end{array} \quad \left\{\begin{array}{l}
a+b=1, \\
ab=3.
\end{array}\right.\right. \text { (*) }
$$

Applying Vieta's theorem, we get
$\left\{\begin{array}{l}a=-1, \\ b=2\end{array}\right.$ or $\left\{\begin{array}{l}a=2, \\ b=-1 .\end{array}\right.$ (*) has no real roots.
$$
\therefore(x+14)(x-16)=-1
$$

or $(x+14)(x-16)=32$,
which gives $x^{2}-2x-223=0$ or $x^{2}-2x-255=0$.
Solving these, we get $x=1 \pm 4 \sqrt{14}, x=1 \pm \sqrt{257}$.
$$
x_{2}=1-4 \sqrt{14}, x_{3}=1+\sqrt{257}, x_{4}=1-\sqrt{257} .
$$