Question

answer this question fast...​

Answer 1

Answer:

Zero

Step-by-step explanation:

x+y=4

Cube on both the sides

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

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

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

Answer 2

$\sf\large\green{\underline { identity \: used....}} \\ \\ \sf\large\green{\underline { }}( {a + b)}^{3} = {a}^{3} + 3 {a}^{2} b + 3a {b}^{2} + {b}^{3}$

$\sf\large\green{\underline { solution : - }} \\ \\ \sf\large\green{\underline { }}(x + y) = 4 \\ \\ \sf\large\green{\underline { }}cubing \: both \: the \: sides... \\ \\ ( {x + y)}^{3} = {4}^{3} \\ \\ \sf\large\green{\underline { }} {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} = 64 \\ \\ \sf\large\green{\underline { }} {x}^{3} + 3xy(x + y) + {y}^{3} = 64 \\ \\ \sf\large\green{\underline { }} {x}^{3} + 3xy(4) + {y}^{3} - 64 = 0 \\ \\ \sf\large\green{\underline { {x}^{3} + 12xy + {y}^{3} - 64 = 0 }} \\ \\ \sf\large\red{ the \:answer \: is \: 0 .}$