Question 13 - End-of-Chapter Exercises - Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I)

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Question 13 (i) Find the value of (i) 𝑥^3+𝑦^3−12𝑥𝑦+64, when 𝑥+𝑦=−4 Our expression is 𝑥^3+𝑦^3−12𝑥𝑦+64 =𝑥^3+𝑦^3+64−12𝑥𝑦 =𝒙^𝟑+𝒚^𝟑+𝟒^𝟑−𝟑 × 𝒙 × 𝒚 × 𝟒 =(𝒙+𝒚+𝟒)(𝒙^𝟐+𝒚^𝟐+𝟒^𝟐−𝒙𝒚−𝟒𝒙−𝟒𝒚) Putting 𝑥+𝑦=−4 Using 𝒙^𝟑+𝒚^𝟑+𝒛^𝟑−𝟑𝒙𝒚𝒛=(𝒙+𝒚+𝒛)(𝒙^𝟐+𝒚^𝟐+𝒛^𝟐−𝒙𝒚−𝒚𝒛−𝒛𝒙) Putting x=𝑥,𝑦=𝑦,𝑧=4 =(−𝟒+𝟒)(𝑥^2+𝑦^2+4^2−𝑥𝑦−4𝑥−4𝑦) =𝟎 × (𝑥^2+𝑦^2+4^2−𝑥𝑦−4𝑥−4𝑦) = 0 Thus, value is 0 Question 13 (ii) Find the value of (ii) 𝑥^3−〖8𝑦〗^3−36𝑥𝑦−216, when 𝑥=2𝑦+6 Our expression is 𝑥^3−〖8𝑦〗^3−36𝑥𝑦−216 =𝑥^3−〖8𝑦〗^3−216−36𝑥𝑦 =𝒙^𝟑+〖(−𝟐𝒚)〗^𝟑 + 〖(−𝟔) 〗^𝟑− 𝟑 × 𝒙 × (−𝟐𝒚) × (−𝟔) =(𝒙−𝟐𝒚−𝟔)(𝒙^𝟐+〖(−𝟐𝒚)〗^𝟐 + 〖(−𝟔)〗^𝟐 − 𝟐𝒙𝒚−𝟔𝒙+𝟏𝟐𝒚) Putting 𝑥=2𝑦+6 Using 𝒙^𝟑+𝒚^𝟑+𝒛^𝟑−𝟑𝒙𝒚𝒛=(𝒙+𝒚+𝒛)(𝒙^𝟐+𝒚^𝟐+𝒛^𝟐−𝒙𝒚−𝒚𝒛−𝒛𝒙) Putting x=𝑥,𝑦=−2𝑦,𝑧=−6 =(𝟐𝒚+𝟔−𝟐𝒚−𝟔)(𝑥^2+〖(−2𝑦)〗^2 + 〖(−6)〗^2 − 2𝑥𝑦−6𝑥+12𝑦) =𝟎 ×(𝑥^2+〖(−2𝑦)〗^2 + 〖(−6)〗^2 − 2𝑥𝑦−6𝑥+12𝑦) = 0 Thus, value is 0