Problem: If (r+1r)2=3\left(r+\dfrac{1}{r}\right)^{2}=3(r+r1)2=3, then r3+1r3r^{3}+\dfrac{1}{r^{3}}r3+r31 equals:
Answer Choices:
A. 111
B. 222
C. 000
D. 333
E. 666 Solution:
r2+1r2=(r+1r)(r2−1+1r2)r^{2}+\dfrac{1}{r^{2}}=\left(r+\dfrac{1}{r}\right)\left(r^{2}-1+\dfrac{1}{r^{2}}\right) r2+r21=(r+r1)(r2−1+r21)
But (r+1r)2=r2+2+1r2=3\text{But } \left(r+\dfrac{1}{r}\right)^{2}=r^{2}+2+\dfrac{1}{r^{2}}=3 But (r+r1)2=r2+2+r21=3
∴r2+1r2=1;∴r2−1+1r2=0;∴r3+1r3=0\therefore r^{2}+\dfrac{1}{r^{2}}=1 ; \quad \therefore r^{2}-1+\dfrac{1}{r^{2}}=0 ; \quad \therefore r^{3}+\dfrac{1}{r^{3}}=0 ∴r2+r21=1;∴r2−1+r21=0;∴r3+r31=0