Find a3−1a3 if a−1a=4
a−1a=4⇒(a−1a)3=(4)3⇒a3−1a3−3a×1a(a−1a)=64⇒a3−1a3−3(4)=64 [∵a−1a=4]⇒a3−1a3−12=64⇒a3−1a3=64+12⇒a3−1a3=76.
Find a3+1a3 if a+1a=5