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

Transcript

Example 14 Factorise the expression 8𝑛^3−60𝑛^2 𝑚+150𝑛𝑚^2−125𝑚^3. Here, There are 2 cube terms: 〖8𝑛〗^3=(2𝑛)^3 and 125𝑚^3=(5𝑚)^3 Since it is −125𝑚^3 is negative, thus we use (a – b)3 Now, 8𝑛^3−60𝑛^2 𝑚+150𝑛𝑚^2−125𝑚^3 =(2𝑛)^3−60𝑛^2 𝑚+150𝑛𝑚^2−(5m)^3 =(𝟐𝒏)^𝟑−𝟑 × (𝟐𝒏)^𝟐 × 𝟓𝐦+𝟑 × 𝟐𝐧 × (𝟓𝐦)^𝟐−(𝟓𝒎)^𝟑 Using (𝑎−𝑏)^3=𝑎^3−3𝑎^2 𝑏+3𝑎𝑏^2−𝑏^3 Putting 𝑎 = 2𝑛, 𝑏 = 5𝑚 = (𝟐𝒏−𝟓𝒎)^𝟑