Question 11 - End-of-Chapter Exercises - Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I)

Transcript

Question 11 If 𝑎+𝑏+𝑐=5 and 𝑎𝑏+𝑏𝑐+𝑐𝑎=10, then prove that 𝑎^3+𝑏^3+𝑐^3−3𝑎𝑏𝑐=−25. Since there are 3 cubes, we use 𝒂^𝟑+𝒃^𝟑+𝒄^𝟑formula Now, 𝒂^𝟑+𝒃^𝟑+𝒄^𝟑−𝟑𝒂𝒃𝒄 =(𝑎+𝑏+𝑐)(𝑎^2+𝑏^2+𝑐^2−𝑎𝑏−𝑏𝑐−𝑎𝑐) Putting 𝑎+𝑏+𝑐=5 =5(𝑎^2+𝑏^2+𝑐^2−[𝑎𝑏+𝑏𝑐+𝑎𝑐]) Putting 𝑎𝑏+𝑏𝑐+𝑐𝑎=10 =𝟓(𝒂^𝟐+𝒃^𝟐+𝒄^𝟐−𝟏𝟎) x3 + y3 + z3 − 75 = 10 × (38 − xy − yz − zx) x3 + y3 + z3 = 10 × (38 − xy − yz − zx) + 75 x3 + y3 + z3 = 10 × [38 − (xy + yz + zx)] + 75 We have to find 𝒂^𝟐+𝒃^𝟐+𝒄^𝟐 To do that, we use the identity (𝒂+𝒃+𝒄)^𝟐=𝒂^𝟐+𝒃^𝟐+𝒄^𝟐+𝟐𝒂𝒃+𝟐𝒃𝒄+𝟐𝒄𝒂 (𝑎+𝑏+𝑐)^2=𝑎^2+𝑏^2+𝑐^2+2 × (𝑎𝑏+𝑏𝑐+𝑐𝑎) Putting 𝑎+𝑏+𝑐=5 and 𝑎𝑏+𝑏𝑐+𝑐𝑎=10 (𝟓)^𝟐=𝒂^𝟐+𝒃^𝟐+𝒄^𝟐+𝟐 × 𝟏𝟎 25=𝑎^2+𝑏^2+𝑐^2+20 25−20=𝑎^2+𝑏^2+𝑐^2 5=𝑎^2+𝑏^2+𝑐^2 𝒂^𝟐+𝒃^𝟐+𝒄^𝟐=𝟓 Putting 𝑎^2+𝑏^2+𝑐^2=5 in (1) 𝒂^𝟑+𝒃^𝟑+𝒄^𝟑−𝟑𝒂𝒃𝒄=5(𝑎^2+𝑏^2+𝑐^2−10) =5(5−10) =5 × −5 =−𝟐𝟓 Hence proved