find value of x3-12xy-64, when x-y-4=0
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x³ +y³ - 12xy + 64
= x³ + y³ + 64 - 3(4xy) => Notice that the polynomial is in the form of:
a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ac)
therefore it factors as:
x³ + y³ + 64 - 12xy
= (x + y + 4)(x² + y² + 16 - xy - 4y - 4x) => substitute for x + y = -4
= (-4 + 4)(x² + y^2 + 16 - xy - 4y - 4x)
= 0 * (x² + y² + 16 - xy - 4y - 4x)
= 0
= x³ + y³ + 64 - 3(4xy) => Notice that the polynomial is in the form of:
a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ac)
therefore it factors as:
x³ + y³ + 64 - 12xy
= (x + y + 4)(x² + y² + 16 - xy - 4y - 4x) => substitute for x + y = -4
= (-4 + 4)(x² + y^2 + 16 - xy - 4y - 4x)
= 0 * (x² + y² + 16 - xy - 4y - 4x)
= 0
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