1.89 If $t=\frac{1}{1-\sqrt[4]{2}}$, then $t$ equals
(A) $(1-\sqrt[4]{2})(2-\sqrt{2})$.
(B) $(1-\sqrt[4]{2})(1+\sqrt{2})$.
(C) $(1+\sqrt[4]{2})(1-\sqrt{2})$.
(D) $(1+\sqrt[4]{2})(1+\sqrt{2})$.
(E) $-(1+\sqrt[4]{2})(1+\sqrt{2})$.
(28th American High School Mathematics Examination, 1977)
$$
\text { [Solution] } \begin{aligned}
\frac{1}{1-\sqrt[4]{2}} & =\frac{1}{1-\sqrt[4]{2}} \cdot \frac{1+\sqrt[4]{2}}{1+\sqrt[4]{2}}=\frac{1+\sqrt[4]{2}}{1-\sqrt{2}} \cdot \frac{1+\sqrt{2}}{1+\sqrt{2}} \\
& =-(1+\sqrt[4]{2})(1+\sqrt{2}) .
\end{aligned}
$$
[Solution] $\frac{1}{1-\sqrt[4]{2}}=\frac{1}{1-\sqrt[4]{2}} \cdot \frac{1+\sqrt[4]{2}}{1+\sqrt[4]{2}}=\frac{1+\sqrt[4]{2}}{1-\sqrt{2}} \cdot \frac{1+\sqrt{2}}{1+\sqrt{2}}$

Therefore, the answer is $(E)$.