Math, asked by pinkikumari97192, 6 hours ago

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

Answers

Answered by faiza96
0

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

Answered by Anonymous
33

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)
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