Ex 9.5, 5 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 7

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Ex 9.5, 5 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 8

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  1. Chapter 9 Class 8 Algebraic Expressions and Identities
  2. Serial order wise

Transcript

Ex 9.5, 5 Show that. (v) (๐‘Žโˆ’๐‘) (๐‘Ž+๐‘)+(๐‘โˆ’๐‘)(๐‘+๐‘)+(๐‘โˆ’๐‘Ž)(๐‘+๐‘Ž)=0 Solving LHS (๐‘Žโˆ’๐‘) (๐‘Ž+๐‘)+(๐‘โˆ’๐‘)(๐‘+๐‘)+(๐‘โˆ’๐‘Ž)(๐‘+๐‘Ž) Using (๐‘ฅโˆ’๐‘ฆ)(๐‘ฅ+๐‘ฆ)=๐‘ฅ^2+๐‘ฆ^2 = (๐‘Ž^2โˆ’๐‘^2 )+(๐‘^2โˆ’๐‘^2 )+(๐‘^2โˆ’๐‘^2 ) = ๐‘Ž^2โˆ’๐‘^2+๐‘^2โˆ’๐‘^2+๐‘^2โˆ’๐‘Ž^2 = (๐‘Ž^2โˆ’๐‘Ž^2 )+(โˆ’๐‘^2+๐‘^2 )+(โˆ’๐‘^2+๐‘^2 ) = 0+0+0 = 0 = R.H.S Since LHS = RHS Hence proved

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.