Problem: If (a+1a)2=3\left(a+\dfrac{1}{a}\right)^{2}=3(a+a1)2=3, then a3+1a3a^{3}+\dfrac{1}{a^{3}}a3+a31 equals:
Answer Choices:
A. 1033\dfrac{10 \sqrt{3}}{3}3103
B. 333 \sqrt{3}33
C. 000
D. 777 \sqrt{7}77
E. 636 \sqrt{3}63 Solution:
a3+1a3=(a+1a)3−3(a+1a)=33−33=0a^{3}+\dfrac{1}{a^{3}}=\left(a+\dfrac{1}{a}\right)^{3}-3\left(a+\dfrac{1}{a}\right)=3 \sqrt{3}-3 \sqrt{3}=0a3+a31=(a+a1)3−3(a+a1)=33−33=0.