Question

x^3 + y^3 - 12xy +64 when x + y = -4

Answer 1

$\large{\boxed{\bold{Answer:}}}$


${x}^{3} + {y}^{3} - 12xy + 64 \\ \\ = > { x}^{3} + {y}^{3} + (4 {)}^{3} - 3(4xy) \\ \\ = > (x + y + 4)( {x}^{2} + {y}^{2} + {4}^{2} - xy - 4y - 4x) \\ \\ = > x + y = - 4 \\ \\ = > ( - 4 + 4)( {x}^{2} + {y}^{2} + {4}^{2} - xy - 4y - 4x) \\ \\ = > (0)( {x}^{2} + {y}^{2} + {4}^{2} - xy - 4y - 4x) \\ \\ = 0$

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Answer 2

Given,

$x + y = -4 \\ \\ \text{Making cube of both sides} \\ \\ (x + y)^{3} = (-4)^{3} \\ \\ x^{3} + y^{3} + 3xy(x+y) = -64 \\ \\ x^{3} + y^{3} + 3xy* (-4) = -64 \\ \\ x^{3} + y^{3} - 12xy = -64 \\ \\ x^{3} + y^{3} - 12xy + 64 = 0 \ \ \textbf{Ans.}$