Question

Using the formula (a-b)^2=(a^2-2ab+b^2), evaluate: i》(196)^2 ii》(689)^2​

Answer 1

GivEn:

• (a - b)² = (a² + b² - 2ab)

To find:

• i》(196)²
• ii》(689)²

Solution:

• let (196)² be (200 - 4)².

Given, Identity - (a - b)² = (a² + b² - 2ab).

Where,

• A = 200
• B = 4

》(196)²

》(200 - 4)²

★ (a - b)² = (a² + b² - 2ab)

》(200)² + (4)² - 2(200)(4)

》40000 + 16 - 1600

》40016 - 1600

》38416

∴ Hence, (196)² = 38416.

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• Let (689)² be (700 - 11)²

Given, Identity - (a - b)² = (a² + b² - 2ab).

Where,

• A = 700
• B = 11

》(689)²

》(700 - 11)²

★ (a - b)² = (a² + b² - 2ab)

》(700)² + (11)² - 2(700)(11)

》490000 + 121 - 15400

》490121 - 15400

》474721

∴ Hence, (689)² = 474721

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⠀⠀⠀⠀⠀⠀》Algebraic identities :

• (a+ b)² = a² + b² + 2ab

• ( a - b )² = a² + b² - 2ab

• ( a + b )² + ( a - b)² = 2a² + 2b²

• ( a + b )² - ( a - b)² = 4ab

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

• a² + b² = ( a + b)² - 2ab

• (a + b )³ = a³ + b³ + 3ab ( a + b)

• ( a - b)³ = a³ - b³ - 3ab ( a - b)

• If a + b + c = 0 then a³ + b³ + c³ = 3abc

Answer 2

Given:

Using the formula (a-b)^2=(a^2-2ab+b^2), evaluate:

1. 196²
2. 689²

Solution:

1. 196²

$\pmb {\tt{\underline{\underline {So, \: According \: to \: the \: question, \: we \: need \: to \: evaluate \: 196² \: by \: using \: the \: formula \: (a-b)^2=(a^2-2ab+b^2)}}}}$

Let 196² be (200 – 4)²

Also Let,

• a = 200

• b = 4

$\\ \\ \\✞ {\tt{\underline{\underline{(a-b)^2 =(a^2-2ab+b^2)}}}} \: \: \: \: \: \: \:\\ \\ \\ \tt \: = {(200 -4) }^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \tt= {200}^{2} - 2(200 \times 4) + {4}^{2} \\ \\ \\ \: \: \: \: \: \: \tt = 40000 - (2 \times 800) + 16 \\ \\ \\ \tt = 40000 - 1600 + 16 \\ \\ \\ \tt = 38400 + 16 \\ \\ \\ \tt \: = 38416$

• Hence, 196² = 38416

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2. 689².

$\pmb {\tt{\underline{\underline{So, \: According \: to \: the \: question, \: we \: need \: to \: evaluate \: 196² \: by \: using \: the \: formula \: (a-b)^2=(a^2-2ab+b^2)}}}}$

Let 689² be (700 – 11)²

Also Let,

• a = 700

• b = 11

$\\ \\ \\✞ {\tt{\underline{\underline{(a-b)^2 =(a^2-2ab+b^2)}}}} \: \: \: \: \: \: \:\\ \\ \\ \tt \: = {(700 -11) }^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \tt= {700}^{2} - 2(700 \times 11) + {11}^{2} \\ \\ \\ \: \: \: \: \: \: \tt = 490000 - (2 \times 7700) + 121 \\ \\ \\ \tt = 40000 - 15400 + 121 \\ \\ \\ \tt = 474600 + 121 \\ \\ \\ \tt \: = 474721$

• Hence, 689² = 474721

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Algebric identities:

1. (a + b)² = a² + 2ab + b²
2. (a - b)² = a² – 2ab + b²
3. a² – b² = (a + b) (a – b)
4. (a + b)³ = a³ + 3a²b + 3ab² + b³
5. (a – b)³ = a³ – 3a²b + 3ab² + b³
6. a³ + b³ = (a + b)(a² – ab + b²)
7. a³ – b³ = (a – b) (a² + ab + b²)