[Class 9] Factorise the expression 8š‘›^3āˆ’60š‘›^2 š‘š+150š‘›š‘š^2āˆ’125š‘š^3 - Finding New Algebraic Identities

part 2 - Example 14 - Finding New Algebraic Identities - Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I) - Class 9

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Transcript

Example 14 Factorise the expression 8š‘›^3āˆ’60š‘›^2 š‘š+150š‘›š‘š^2āˆ’125š‘š^3. Here, There are 2 cube terms: 怖8š‘›ć€—^3=(2š‘›)^3 and 125š‘š^3=(5š‘š)^3 Since it is āˆ’125š‘š^3 is negative, thus we use (a – b)3 Now, 8š‘›^3āˆ’60š‘›^2 š‘š+150š‘›š‘š^2āˆ’125š‘š^3 =(2š‘›)^3āˆ’60š‘›^2 š‘š+150š‘›š‘š^2āˆ’(5m)^3 =(šŸš’)^šŸ‘āˆ’šŸ‘ Ɨ (šŸš’)^šŸ Ɨ šŸ“š¦+šŸ‘ Ɨ šŸš§ Ɨ (šŸ“š¦)^šŸāˆ’(šŸ“š’Ž)^šŸ‘ Using (š‘Žāˆ’š‘)^3=š‘Ž^3āˆ’3š‘Ž^2 š‘+3š‘Žš‘^2āˆ’š‘^3 Putting š‘Ž = 2š‘›, š‘ = 5š‘š = (šŸš’āˆ’šŸ“š’Ž)^šŸ‘

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