6 th no. question. answer = a
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Given x3+ 1/x3 =110
Recall that (x+1/x)3 = x3+ 1/x3 + 3(x+1/x)
Let (x+1/x)=a
⇒ a3 = 110 + 3a
⇒a3 − 3a −110=0
Put a=5
⇒ 53 − 3(5) −110
=125 −15 −110=0
Hence (a − 5) is a factor.
On dividing (a3 − 3a −110) with (a − 5) we get the quotient as (a2+5a+22)
∴ a3 − 3a −110 = (a − 5)(a2+5a+22) = 0
Hence (a − 5) = 0 and (a2+5a+22) ≠ 0
⇒(x+1/x) − 5=0
∴ (x+1/x) = 5
Recall that (x+1/x)3 = x3+ 1/x3 + 3(x+1/x)
Let (x+1/x)=a
⇒ a3 = 110 + 3a
⇒a3 − 3a −110=0
Put a=5
⇒ 53 − 3(5) −110
=125 −15 −110=0
Hence (a − 5) is a factor.
On dividing (a3 − 3a −110) with (a − 5) we get the quotient as (a2+5a+22)
∴ a3 − 3a −110 = (a − 5)(a2+5a+22) = 0
Hence (a − 5) = 0 and (a2+5a+22) ≠ 0
⇒(x+1/x) − 5=0
∴ (x+1/x) = 5
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