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Question
Factorize `[x^2 + 1/x^2] - 4[x + 1/x] + 6`
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Solution
`= x^2 + 1/x^2 - 4x - 4/x + 4 + 2`
`= x^2 + 1/x^2 + 4 + 2 - 4/x - 4x`
`= (x^2) + (1/x)^2 + (-2)^2 + 2 xx x xx x 1/x + 2x 1/x xx (-2) + 2(-2)x`
Using identity
a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2
We get
`= [x + 1/x + (-2)]^2`
`= [x + 1/x - 2]^2`
`= [x + 1/x - 2][x + 1/x - 2]`
`∴ [x^2 + 1/x^2] - 4[x + 1/x] + 6 = [x + 1/x - 2][x + 1/x - 2]`
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