Question

x+y+1=0 prove that x³+y³+1³=3xy

Answer 1

x+y= -1
x³+y³+3xy(x+y)= (-1)³
x³+y³-3xy= -1
x³+y³+1³=3xy
[ Proved ]
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Answer 2

x³+y³+z³ - 3xy = (x + y + z)(x² + y² + z² - ab - ac - bc)

Apply this equation on it

x³+y³+1³ - 3xy = (x + y + 1)(x² + y² + z² - ab - ac - bc)

if x + y + 1 = 0

x³+y³+1³ - 3xy = 0

x³+y³+1³ = 3xy