a+1\a=p prove that a3+1\a3=p(p2-3) a3 is a cube and p2 is p square solve it fast plszzz
kundan7452:
a+1/a=p,.....then {a+1/a}^3 =p^3 ,....i.e.a^3+1/a^3+3×a×1/a(a+1/a)=p^3,.....after solving this we get p(p^2-3)=a^3+1/a^3
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If a+1\a=p then, prove that a3+1\a3=p(p2-3)
Taking RHS
p(p2-3)
Given that p=a+1/a
= a+1/a[(a+1/a)2-3]
= a+1/a(a2+1/a2+2-3]
= a[a2+1/a2-1]+1/a[a2+1/a2-1]
= a3+1/a-a+a+1/a3-1/a
= a3+1/a3-a+a+1/a-1/a
= a3+1/a3
This is the LHS
Hence proved that RHS = LHS
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