X^3+y^3+15xy-125 factorise
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X^3+y^3+15xy-125
(x)^3 +(y)^3 +(-5)^3 - (3)(x)(y)(-5)
now, it is in the form of
a^3 + b^3 + c^3 - 3abc
use this identity :
a^3 + b^3 + c^3 - 3abc = (a +b+c)(a^2 +b^2 + c^2 -ab -bc -ca)
which equals
{x + y + (-5)} {x^2 + y^2 + 25 - (x)(y) -(y)(-5) -(-5)(x)}
= (x + y -5)( x^2 + y^2 + 25 -xy + 5y + 5x)
(x)^3 +(y)^3 +(-5)^3 - (3)(x)(y)(-5)
now, it is in the form of
a^3 + b^3 + c^3 - 3abc
use this identity :
a^3 + b^3 + c^3 - 3abc = (a +b+c)(a^2 +b^2 + c^2 -ab -bc -ca)
which equals
{x + y + (-5)} {x^2 + y^2 + 25 - (x)(y) -(y)(-5) -(-5)(x)}
= (x + y -5)( x^2 + y^2 + 25 -xy + 5y + 5x)
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