Question

find the value of x ^3 + y ^3+ 12 x y - 64 when X + Y =4​

Answer 1

Answer:

0

Step-by-step explanation:

$x + y = 4$

${(x + y)}^{3} = 64$

${x}^{3} + {y}^{3} + 3xy(x + y) = 64$

${x}^{3} + {y}^{3} + 3xy(4) = 64$

${x}^{3} + {y}^{3} + 12xy - 64 = 0$