Question

if x+y=1+xy , show that x3+y3=1+x3y3​

Answer 1

Required Answer:-

Given:

• x + y = 1 + xy

To Prove:

• x³ + y³ = 1 + x³y³

Proof:

We have,

➡ x + y = 1 + xy

Cubing both sides, we get,

➡ (x + y)³ = (1 + xy)³

➡ x³ + y³ + 3xy(x + y) = 1 + x³y³ + 3xy(1 + xy)

Shifting 3xy(x + y) on right side, we get,

➡ x³ + y³ = 1 + x³y³ + 3xy(1 + xy) - 3xy(x + y)

➡ x³ + y³ = 1 + x³y³ + 3xy(1 + xy - x - y)

As x + y = 1 + xy,

➡ x³ + y³ = 1 + x³y³ + 3xy(x + y - x - y)

➡ x³ + y³ = 1 + x³y³ + 3xy × 0

➡ x³ + y³ = 1 + x³y³ (Hence Proved)

Formula Used:

• (a + b)³ = a³ + 3ab(a + b) + b³

More Formulae To Know:

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• a² - b² = (a + b)(a - b)
• (a - b)³ = a³ - 3ab(a - b) - b³