Question

If a+b=5 and a²+b²=11 , then prove that a³+b³=20

Answer 1

$\huge\mathcal{Heya}$
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Given :-

a + b = 5

a² + b² = 11

To prove :-

a³ + b³ = 20

Solution :-

Using identity

(a + b)² = a² + b² + 2ab

Putting given values in it.

We get

(5)² = 11 + 2ab

25 = 11 + 2ab

2ab = 25 - 11

2ab = 14

ab = 14 / 2

ab = 7

Now,

Using another identity

a³ + b³ = (a + b)(a² + b² - ab)

R.H.S. = (a + b)(a² + b² - ab)

Putting given values in it.

We get

= (5)(11 - 7)

= 5 × 4

= 20

= L.H.S.

Hence, proved.

This kind of questions can be solved by using identities.

Some other identities are :-

(a - b)² = a² + b² - 2ab

(a - b)³ = a³ - b³ - 3a²b + 3ab²

(a + b)³ = a³ + b³ + 3a²b + 3ab²

a³ - b³ = (a - b)(a² + b² + ab)

a² - b² = (a + b)(a - b)

$\huge\mathbb{Hope \ this \ helps.}$

Answer 2

Step-by-step explanation:

Given :-

a + b = 5

a² + b² = 11

To prove :-

a³ + b³ = 20

Solution :-

Using identity

(a + b)² = a² + b² + 2ab

Putting given values in it.

We get

(5)² = 11 + 2ab

25 = 11 + 2ab

2ab = 25 - 11

2ab = 14

ab = 14 / 2

ab = 7

Now,

Using another identity

a³ + b³ = (a + b)(a² + b² - ab)

R.H.S. = (a + b)(a² + b² - ab)

Putting given values in it.

We get

= (5)(11 - 7)

= 5 × 4

= 20

= L.H.S.

Hence, proved.