Question

a+b=11 and a^2+b^2=65 ; find a^3+b^3


ans 407​

Answer 1

Given : a + b = 11, a² + b² = 65

To find : a³ + b³

Solution :

a² + b² = 65

⇒ (a + b)² - 2ab = 65

⇒ 11² - 2ab = 65

⇒ 2ab = 121 - 65

⇒ 2ab = 56

⇒ ab = 28

Now, a³ + b³

= (a + b) (a² + b² - ab)

= 11 * (65 - 28)

= 11 * 37

= 407

⇒ a³ + b³ = 407

Identity rules :

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• (a + b)³ = a³ + 3a²b + 3ab² + b³
• (a - b)³ = a³ - 3a²b + 3ab² - b³
• a³ + b³ = (a + b) (a² - ab + b²)
• a³ - b³ = (a - b) (a² + ab + b²)

Answer 2

Solution:

It is given that,

a+b = 11 ---(1)

a²+b² = 65 ----(2)

we know the algebraic identity:

a²+b²+2ab = (a+b)²

=> 65 + 2ab = (11)²

/* from (1) & (2) */

=> 65 + 2ab = 121

=> 2ab = 121-65

=> 2ab = 56

=> ab = 56/2 = 28 ---(3)

Now ,

a³+b³ = (a+b)³ - 3ab(a+b)

= (11)³ - 3×28×11

= 1331 - 924

= 407

Therefore,

a³+b³ = 407

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