Ex 2.5,9 - Chapter 2 Class 9 Polynomials
Last updated at July 6, 2018 by Teachoo
Transcript
Ex 2.5, 9 Verify: (i) x3 + y3 = (x + y) (x2 – xy + y2) L.H.S x3 + y3 We know (x + y)3 = x3 + y3 + 3xy (x + y) So, x3 + y3 = (x + y)3 – 3xy (x + y) = (x + y)3 – 3xy (x + y) = (x + y) [ (x + y)2 – 3xy)] Using (a+b)2 = a2 + b2 + 2ab = (x + y) [(x2 + y2 + 2xy) – 3xy] = (x + y) (x2 + y2 – xy) = (x + y) (x2 - xy + y2) = R.H.S R.H.S (x + y) (x2 – xy + y2) Hence proved Ex 2.5, 9 Verify: (ii) x3 - y3 = (x - y) (x2 + xy + y2) L.H.S x3 - y3 We know (x - y)3 = x3 - y3 - 3xy (x - y) x3 - y3= (x - y)3 +3xy (x - y) = (x - y)3 + 3xy (x - y) = (x - y) [ (x - y)2 + 3xy)] = (x - y) [(x2 + y2 - 2xy) + 3xy] = (x - y) (x2 + y2 + xy) = (x - y) (x2 + xy + y2) = R.H.S R.H.S (x - y) (x2 + xy + y2) Hence proved
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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.
