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Sir pls solve the following.
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Asked by rsudipto | 04 Jan, 2019, 19:17: PM
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begin mathsize 16px style Consider comma open parentheses straight r subscript 1 plus straight r subscript 2 plus straight r subscript 3 minus straight r close parentheses squared equals straight r subscript 1 superscript 2 plus straight r subscript 2 superscript 2 plus straight r subscript 3 superscript 2 plus straight r squared minus 2 straight r open parentheses straight r subscript 1 plus straight r subscript 2 plus straight r subscript 3 close parentheses plus 2 open parentheses straight r subscript 1 straight r subscript 2 plus straight r subscript 2 straight r subscript 3 plus straight r subscript 3 straight r subscript 1 close parentheses straight r subscript 1 plus straight r subscript 2 plus straight r subscript 3 minus straight r equals 4 straight R space and space straight r subscript 1 straight r subscript 2 plus straight r subscript 2 straight r subscript 3 plus straight r subscript 3 straight r subscript 1 equals straight s squared rightwards double arrow open parentheses 4 straight R close parentheses squared equals straight r subscript 1 superscript 2 plus straight r subscript 2 superscript 2 plus straight r subscript 3 superscript 2 plus straight r squared minus 2 straight r open parentheses straight r subscript 1 plus straight r subscript 2 plus straight r subscript 3 close parentheses plus 2 straight s squared rightwards double arrow straight r subscript 1 superscript 2 plus straight r subscript 2 superscript 2 plus straight r subscript 3 superscript 2 plus straight r squared equals 16 straight R squared plus 2 straight r open parentheses straight r subscript 1 plus straight r subscript 2 plus straight r subscript 3 close parentheses minus 2 straight s squared rightwards double arrow 2 open parentheses rr subscript 1 plus rr subscript 2 plus rr subscript 3 close parentheses equals 2 open square brackets fraction numerator straight s squared over denominator straight s open parentheses straight s minus straight a close parentheses end fraction plus fraction numerator straight s squared over denominator straight s open parentheses straight s minus straight b close parentheses end fraction plus fraction numerator straight s squared over denominator straight s open parentheses straight s minus straight c close parentheses end fraction close square brackets Simplifying space this space we space get straight r subscript 1 superscript 2 plus straight r subscript 2 superscript 2 plus straight r subscript 3 superscript 2 plus straight r squared equals 16 straight R squared minus open parentheses straight a squared plus straight b squared plus straight c squared close parentheses straight r subscript 1 superscript 2 plus straight r subscript 2 superscript 2 plus straight r subscript 3 superscript 2 plus straight r squared plus straight a squared plus straight b squared plus straight c squared equals 16 straight R squared end style
Answered by Sneha shidid | 07 Jan, 2019, 10:11: AM
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