Question

if a+b=10 and a^2+b^2=58.find a^3+b^3 solve with steps​

Answer 1

Answer:So the answer is Given a+b =10

so  (a+b)

2

=a

2

+b

2

+2ab

10

2

=58+2ab

∴2ab=100−58=42

∴ab=21

(a+b)

3

=a

3

+b

3

+3ab(a+b)

10

3

=a

3

+b

3

+3×21×10

∴a

3

+b

3

=1000−630=370

Step-by-step explanation:

Answer 2

Answer:

Given:

a + b = 10 and

${a}^{2} + b {}^{2} = 58$

To find:

${a}^{3} + b {}^{3}$

Solution:

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

$= (10)( {a}^{2} + {b}^{2} - ab) \\ \\ = (10)(58 - ab)$

In order to find ab

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

$\Longrightarrow2ab = {a}^{2} + b {}^{2} \\ \\ \Longrightarrow2ab = 100 - 58 = 42$

Thus, ab = 21

So, (10)(58 – ab) = 10(58 – 21)

= 10 × 37

= 370

Hence, a^3 + b^3 = 370.

Some algebraic identities:

→ (a + b)^2 = a^2 + 2ab + b^2

→ (a – b)^2 = a^2 – 2ab + b^2

→ a^2 – b^2 = (a + b) (a – b)

→ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

→ (a + b – c)^2 = a^2 + b^2 + c^2 + 2ab – 2bc – 2ca

→ (a – b – c)^2 = a^2 + b^2 + c^2 – 2ab + 2bc – 2ca

→ (a + b)^3 = a^3 + b^3 + 3ab(a + b)

→ (a – b)^3 = a^3 – b^3 – 3ab(a – b)

→ (a^3 + b^3) = (a + b) (a^2 – ab + b^2)

→ (a^3 – b^3) = (a – b) (a^2 + ab + b^2)