Problem: If t=11−24t=\dfrac{1}{1-\sqrt[4]{2}}t=1−421, then ttt equals
Answer Choices:
A. (1−24)(2−2)(1-\sqrt[4]{2})(2-\sqrt{2})(1−42)(2−2)
B. (1−24)(1+2)(1-\sqrt[4]{2})(1+\sqrt{2})(1−42)(1+2)
C. (1+24)(1−2)(1+\sqrt[4]{2})(1-\sqrt{2})(1+42)(1−2)
D. (1+24)(1+2)(1+\sqrt[4]{2})(1+\sqrt{2})(1+42)(1+2)
E. −(1+24)(1+2)-(1+\sqrt[4]{2})(1+\sqrt{2})−(1+42)(1+2) Solution:
11−24=11−24⋅1+241+24=1+241−2⋅1+21+2=−(1+24)(1+2)\dfrac{1}{1-\sqrt[4]{2}}=\dfrac{1}{1-\sqrt[4]{2}} \cdot \dfrac{1+\sqrt[4]{2}}{1+\sqrt[4]{2}}=\dfrac{1+\sqrt[4]{2}}{1-\sqrt{2}} \cdot \dfrac{1+\sqrt{2}}{1+\sqrt{2}}=-(1+\sqrt[4]{2})(1+\sqrt{2})1−421=1−421⋅1+421+42=1−21+42⋅1+21+2=−(1+42)(1+2).