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Algebra Identities and Formulas
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Last updated at May 29, 2023 by Teachoo
Question 5 Show that. (v) (𝑎−𝑏) (𝑎+𝑏)+(𝑏−𝑐)(𝑏+𝑐)+(𝑐−𝑎)(𝑐+𝑎)=0 Solving LHS (𝑎−𝑏) (𝑎+𝑏)+(𝑏−𝑐)(𝑏+𝑐)+(𝑐−𝑎)(𝑐+𝑎) Using (𝑥−𝑦)(𝑥+𝑦)=𝑥^2+𝑦^2 = (𝑎^2−𝑏^2 )+(𝑏^2−𝑐^2 )+(𝑏^2−𝑐^2 ) = 𝑎^2−𝑏^2+𝑏^2−𝑐^2+𝑐^2−𝑎^2 = (𝑎^2−𝑎^2 )+(−𝑏^2+𝑏^2 )+(−𝑐^2+𝑐^2 ) = 0+0+0 = 0 = R.H.S Since LHS = RHS Hence proved