If a³+1/a³=18 then a²+1/a²=?
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a³+b³= (a+b)(a²+b²-ab)
a³+1/a³
= (a+1/a)(a²+1/a²-a*1/a)
= (a+1/a){(a+1/a)² -3}
let a+1/a =t
thus
t(t²-3) = 18
t³-3t-18= 0
by hit and trial
method we got its one root as 3
thus
t³-3t-18= (t-3)(t²+3t+6)
this cubic have only one +ve root
that s why t = 3
a+1/a = 3
squaring both sides
a²+1/a²+2 = 9
a²+1/a² = 7
thus 7 will be answer
a³+1/a³
= (a+1/a)(a²+1/a²-a*1/a)
= (a+1/a){(a+1/a)² -3}
let a+1/a =t
thus
t(t²-3) = 18
t³-3t-18= 0
by hit and trial
method we got its one root as 3
thus
t³-3t-18= (t-3)(t²+3t+6)
this cubic have only one +ve root
that s why t = 3
a+1/a = 3
squaring both sides
a²+1/a²+2 = 9
a²+1/a² = 7
thus 7 will be answer
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