Question

The factorisation of 64p² – 81q² involves :

a) (a + b)² = a² + 2ab + b²

b) (a – b)² = a² – 2ab + b²

c) (a + b)(a – b) = a² + b²

d) None of these

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Answer 1

Question :

The factorization of 64p² – 81q² involves :

a. (a + b)² = a² + 2ab + b²

b. (a – b)² = a² – 2ab + b²

c. (a + b)(a – b) = a² – b²

d. None of these

Answer :

c. (a + b)(a – b) = a² + b²

Explanation :

64p² - 81q²

⇒ (8p)² - (9q)²

∵ a² - b² = (a+b) (a-b)

⇒ ( 8p + 9q ) ( 8p - 9q )

More Info :

• ( a + b )³ = a³ + b³ + 3ab ( a + b )
• ( a - b )³ = a³ - b³ - 3ab ( a - b )
• a³ + b³ = ( a + b ) ( a² - ab + b² )
• a³ - b³ = ( a - b ) ( a² + ab + b² )

Answer 2

Answer:

$\pink\bigstar$Option (C) (a + b) (a - b) = a² - b²

Step-by-step explanation:

↪ 64p² - 81q²

↪ (8p)² - (9q)²

[ °.° a² - b² = (a + b) (a - b) ]

↪ (8p + 9q) (8p - 9q)

$\red{\bigstar}$ Some algebraic identities. $\green{\bigstar}$

•(a + b)² = a² + b² + 2ab

•(a - b)² = a² + b² - 2ab

•(a + b)(a – b) = a² - b²

•(a + b)³ = a³ + b³ + 3ab(a + b)

•(a - b)³ = a³ - b³ + 3ab(a + b)

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