Question

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

Answer 1

Step-by-step explanation:

We have (a – b)2 = (a2 – 2ab + b2)(689)2 can be written as 700-11 So here a=700 and b=11 Using the formula, (700 – 11)2 = (7002 – 2 X 700 X 11 + 112) On simplifying we get (700 – 11)2 = (490000 – 15400 +121) (689)2 = 474721Read more on Sarthaks.com - https://www.sarthaks.com/729549/using-the-formula-a-b-2-a-2-2ab-b-2-evaluate-689-2?show=729562#a729562

Answer 2

Given:

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

1. 689²

Solution:

1. 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)