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x + 1/x = 3 x² + 1 = 3x On squaring, we get (x² + 1)² = 9x² x⁴ + 1 + 2 x² = 9 x² x⁴ + 1 = 7x² Now, `(x^3 (x+3)+x (5x+3)+1)/(x^4+1)` ⇒ `[(x^4 + 3x)^3 +(5x^2 + 3x + 1)]/(x^4 + 1)` ⇒ `[x^(4 + 1) + 5x^(2) + 3x (x)^2 + 1]/x^(4 + 1)` ⇒ `(7x^2 + 5x^2 + 3x(3x))/(7x^2)` ⇒ `(7x^2+5x^2+9x^2)/(7x^2)` `(21x^2)/(7x^2)` = 3
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