Question

if a+b=11,ab=28 find a³+b³​

Answer 1

$\bf\huge\underline{Answer}$

The value of a³ + b³ is 407 respectively.

Solution :-

Given :

a + b = 11

ab = 28

We know that,

${a}^{3} + {b}^{3} = {(a + b)}^{3} - 3ab(a + b)$

Putting the values.

$= > {a}^{3} + {b}^{3} = {(11)}^{3} - 3 \times 28(11)$

$= > {a}^{3} + {b}^{3} = 1331 - 3 \times 308$

$= > {a}^{3} + {b}^{3} = 1331 - 924$

$= > {a}^{3} + {b}^{3} = 407$

Hence, the value of a³ + b³ is 407

Identity used :-

• a³ + b³ = (a + b)³ - 3ab(a + b)

Answer 2

Answer :-

a³ + b³ = 407

Solution :-

We know that

(a + b)³ = a³ + b³ + 3ab(a + b)

Here

• a + b = 11

• ab = 28

By substituting the values

⇒ (11)³ = a³ + b³ + 3(28)(11)

⇒ 1331 = a³ + b³ + 84(11)

⇒ 1331 = a³ + b³ + 924

Transpose 924 to LHS

⇒ 1331 - 924 = a³ + b³

⇒ 407 = a³ + b³

⇒ a³ + b³ = 407

Verification :-

(a + b)³ = a³ + b³ + 3ab(a + b)

⇒ (11)³ = 407 + 3(28)(11)

⇒ 1331 = 407 + 84(11)

⇒ 1331 = 407 + 924

⇒ 1331 = 1331

Identity used :-

• (a + b)³ = a³ + b³ + 3ab(a + b)