Question

If x+ y-1=0,prove that x^3+y^3+3xy=1​

Answer 1

x+y-1=0

x+y=1

x^3+3x^2*y+3x*y^2+y^3=1

x^3+3x*y(x+y)+y^3=1

x^3+3x*y(1)+y^3=1  [since x+y=1]

x^3+3x*y+y^3=1

hence proved

thank you

Answer 2

Note:

★ (a + b)² = a² + 2ab + b²

★ (a - b)² = a² - 2ab + b²

★ a² - b² = (a + b)(a - b)

★ (a + b)³ = a³ + b³ + 3ab(a + b)

★ (a - b)³ = a³ - b³ - 3ab(a - b)

★ a³ + b³ = (a + b)(a² - ab + b²)

★ a³ - b³ = (a - b)(a² + ab + b²)

Solution:

Given : x + y - 1 = 0

To prove : x³ + y³ + 3xy = 1

Proof :

We have ;

=> x + y - 1 = 0

=> x + y = 1 -----------(1)

Now,

Cubing both sides of eq-(1) , we get ;

=> (x + y)³ = 1³

=> x³ + y³ + 3xy(x + y) = 1

=> x³ + y³ + 3xy•1 = 1 { using eq-(1) }

=> x³ + y³ + 3xy = 1

Hence proved .