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The factors of x^{3} − 1 + y^{3} + 3xy are

#### Options

(x − 1 + y) (x

^{2}+ 1 + y^{2}+ x + y − xy)(x + y + 1) (x

^{2}+ y^{2}+ 1 −xy − x − y)(x − 1 + y) (x

^{2}− 1 − y^{2}+ x + y + xy)3(x + y −1) (x

^{2}+ y^{2}− 1)

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#### Solution

The given expression to be factorized is x^{3} − 1 + y^{3} + 3xy

This can be written in the form

x^{3} − 1 + y^{3} + 3xy = `(x)^2 + (-1)^3 + (y)^3 -3 .(x).(-1).(y)`

Recall the formula `a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)`

Using the above formula, we have

x^{3} − 1 + y^{3} + 3xy

` = {x+(-1)+ y}{(x)^2 + (-1)^2 + (y)^2 - (x).(-1) - (-1). (y) - (y).(x)}`

` = (x-1 + y)(x^2 + 1 + y^2 + x+ y -xy)`

So, the correct choice is (a).

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