Question

if a+b=11 and a2+b2=65 find a3+b3​

Answer 1

ANSWER :–

• The value of $\sf {a}^{3} + {b}^{3} = 407$

GIVEN :–

• a + b = 11.
• a² + b² = 65.

TO FIND :–

• $\sf {a}^{3} + {b}^{3}$

SOLUTION :–

Identity used,

$\mapsto {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

And here we have,

• a + b = 11.
• a² + b² = 65.

Now, substitute the given values in the identity.

$\longmapsto{(11)}^{2} = 65 + 2ab$

$\longmapsto121 - 65 = 2ab$

$\longmapsto56 = 2ab$

$\longmapsto\cancel \dfrac{56}{2} = ab$

$\longmapsto 28 = ab$

We get,

• ab = 28.

Now,

$\mapsto {a}^{3} + {b}^{3} = (a + b)( {a}^{2} + {b}^{2} - ab)$

Now we have,

• a + b = 11.
• a² + b² = 65.
• ab = 28.

$\longmapsto {a}^{3} + {b}^{3} = (11)(65 - 28)$

$\longmapsto{a}^{3} + {b}^{3} = 11 \times 37$

$\longmapsto{a}^{3} + {b}^{3} = 407$

Hence,

The value of $\sf {a}^{3} + {b}^{3}$ is 407.

Answer 2

Answer:

1331

Step-by-step explanation:

a+b=11

A^2 +B^2 = 65

To get 11 we have so many options but here we will take it as 7+4 because 7^2 + 4^2 = 65

So A=7,,B=4

Therefore (a+b)^3= a^3 + b^3 + 3a^2b+ 3ab^2

After entering the values,,

(a+b)^3= 7^3 + 4^3 + 3 * 7^2 * 4 + 3 * 7 * 4^2

          = 343 + 64 + 588 + 336

          = 1331