[tex]\large\underline{\sf{Given- }}[/tex]

[tex]\sf \: x + y = 4 \\ \\ \\ [/tex]

[tex]\large\underline{\sf{To\:Find - }}[/tex]

[tex]\sf \: {x}^{3} + {y}^{3} + 12xy - 64 \\ \\ \\ [/tex]

[tex]\large\underline{\sf{Solution-}}[/tex]

Given that,

[tex]\sf \: x + y = 4 \\ \\ [/tex]

can be rewritten as

[tex]\sf \: x + y - 4= 0 \\ \\ [/tex]

[tex]\sf \: x + y + ( - 4)= 0 \\ \\ [/tex]

We know,

[tex]\boxed{ \sf{ \:x + y + z = 0 \: \sf \:  \implies \: {x}^{3} + {y}^{3} + {z}^{3} = 3xyz \: }} \\ \\ [/tex]

So, using this result, we get

[tex]\sf \: {(x)}^{3} + {(y)}^{3} + {( - 4)}^{3} = 3(x)(y)( - 4) \\ \\ [/tex]

[tex]\sf \: {x}^{3} + {y}^{3} - 64 = - 12xy\\ \\ [/tex]

[tex]\sf \: \sf \:  \implies \: {x}^{3} + {y}^{3} + 12xy - 64 = 0\\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]