Question

If x+y+1=0,prove that x3+y3+1=3xy

Answer 1

x+y+1=0
So, x+y = -1
Cubing both side , we get
(x + y ) ^3 = -1
x^3 + y^3 + 3xy (x+y) = -1
x^3 + y^3 + 3xy (-1) ● see the second line● = -1
x^3 + y^3 +1 -3xy =0
So , x^3 + y^3 +1 = 3xy
Hence proved.

Answer 2

Hey......!!! here is ur answer....☺️☺️☺️

Here.... Given that,

x+y+1 = 0

=> x+y = –1......(1)

On squaring equation (1) both the sides...

=> (x+y)² = 1

=> x²+2xy+y² = 1

=> x²+y² = 1–2xy......(2)

Now, we have to prove that......

x³+y³+1 = 3xy

On taking L.H.S.

x³+y³+1

=> (x+y)(x²–xy+y²)+1

{ Because a³+b³ = (a+b)(a²–ab+b²) }

=>(x+y)(x²+y²–xy)+1

=> (–1)(1–2xy–xy)+1 { From equation (1) & (2) }

=>–1(1–3xy)+1

=> –1+3xy+1

=> 3xy = R.H.S.

I hope it will help you......✌️✌️✌️