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Question
Factorise:
`y^2 + (1)/(4y^2) + 1 - 6y - (3)/y`
Sum
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Solution
`y^2 + (1)/(4y^2) + 1 - 6y - (3)/y`
= `(y^2 + 1/(4y^2) + 1) - (6y + 3/y)`
= `(y + 1/(2y))^2 - 6 (y + 1/(2y))`
= `(y + 1/(2y))(y + 1/(2y) - 6)`.
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