Question

write the identity of a³+b³​

Answer 1

Answer:

Here, volume is in cubic units. Hence, a³-b³ = (a-b) (a²+ab+b²). The algebraic identity a³-b³ = (a-b) (a²+ab+b²) has been verified.

Step-by-step explanation:

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Answer 2

Identities :-

• (a+b)² = a² + 2ab + b²
• (a-b)² = a² - 2ab + b²
• (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca
• (x+a)(x+b) = x² + x(a+b) + ab
• a²-b² = (a+b)(a-b)
• (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)
• a³+b³+c³ - 3abc = (a+b+c)(a² + b² + c² - ab - bc - ca)

Conditional identity:-

if a + b + c = 0,

then,

• a³ + b³ + c³ = 3abc

Examples :-

(a+b)² = a² + 2ab + b²

• a = 2
• b = 1

Substituting,

=≥ (2)² + 2(2)(1) + 1²

=≥ 4 + 4 + 1

=≥ 9

(a-b)² = a² - 2ab + b²

• a = 2
• b = 1

Substituting,

=≥ (2)² - 2(2)(1) + (1)²

=≥ 4 - 4 + 1

=≥ 1

(a+b+c)² = a² + b² + c² + 2(ab + bc + ca)

• a = 2
• b = 1
• c = 3

Substituting,

=≥ (2)² + (1)² + (3)² + 2[(2)(1) + (1)(3) + (3)(2)]

=≥ 4 + 1 + 9 + 2(2 + 3 + 6)

=≥ 14 + 2(11)

=≥ 14 + 22

=≥ 36

(x+a)(x+b) = x² + x(a+b) + ab

• a = 2
• b = 1
• x = 3

Substituting,

=≥ (3)² + 3(1+2) + (1)(2)

=≥ 9 + 3(3) + 2

=≥ 9 + 9 + 2

=≥ 20

a²-b² = (a+b)(a-b)

• a = 2
• b = 1

Substituting,

=≥ (2-1)(2+1)

=≥ (1)(3)

=≥ 3

(a+b)³ = a³ + 3ab(a+b) + b³

• a = 2
• b = 1

Substituting,

=≥ (2)³ + 3(2)(1)(2+1) + (1)³

=≥ 8 + 6(3) + 1

=≥ 8 + 18 + 1

=≥ 27

(a-b)³ = a³ - 3ab(a-b) - b³

• a = 2
• b = 1

Substituting,

=≥ (2)³ - 3(2)(1)(2-1) - (1)³

=≥ 8 - 6(1) - 1

=≥ 8 - 6 - 1

=≥ 1

a³+b³ = (a+b)(a² - ab + b²)

• a = 2
• b = 1

Substituting,

=≥ 2³ + 1³

=≥ (2+1)(2² - (2)(1) + 1²)

=≥ (3)(4 - 2 + 1)

=≥ 3(3)

=≥ 9

a³-b³ = (a-b)(a² + ab + b²)

• a = 2
• b = 1

Substituting,

=≥ 2³ - 1³

=≥ (2-1)(2² + (2)(1) + 1²)

=≥ (1)(4 + 2 + 1)

=≥ (1)(7)

=≥ 7

a³+b³+c³ - 3abc = (a+b+c)(a² + b² + c² - ab - bc - ca)

• a = 2
• b = 1
• c = 3

Substituting,

=≥ (2)³ + (1)³ + (3)³ - 3(2)(1)(3)

=≥ (2+1+3)[2² + 1² + 3² - (2)(1) - (1)(3) - (3)(2)]

=≥ (6)(4 + 1 + 9 - 2 - 3 - 6)

=≥ 6(14 - 11)

=≥ 6(3)

=≥ 18

Conditional identity :-

if a + b + c = 0,

then

a³ + b³ + c³ = 3abc

• a = 2
• b = 1
• c = (-3)

Substituting,

=≥ 2 + 1 + (-3)

=≥ 3 + (-3)

=≥ 0

Hence,

=≥ (2)³ + (1)³ + (-3)³

=≥ 3(2)(1)(-3)

=≥ (-18)