Question

if x-y=4and xy=21,then find x3-y3​

Answer 1

Answer:88

Step-by-step explanation:

Given x-y=4

Cube on both sides

(x-y)^3=(4)^3

x^3-3xy (x-y)-y^3=64

x^3-3 (2)(4)-y^3=64

x^3-24-y^3=64

x^3-y^3=64+24

x^3-y^3=88

Answer 2

Answer:

$x - y = 4 \\ xy = 21 \\ \\ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y) \\ {x}^{3} - {y}^{3} = {4}^{3} + (3 \times 21)(4) \\ {x}^{3} - {y}^{3} = 64 + (63 \times 4) \\ {x}^{3} - {y}^{3} = 64 + 252 \\ {x}^{3} - {y}^{3} = \underline{ \underline{ 316}}$