In this meritnation solution of factorisation of ncert book of the fifth solution how did third step come.
Please explain these.

In this meritnation solution of factorisation of ncert book of the fifth solution how did third step come. Please explain these. 21:36 ' 0.04KJs g Page 224 v Solution: [x2 - (x - z)2] [x2+ (x - z)2] - [x - (x - z)] [x + (x - z)] [x2 + (x - z(2x - z) [x2 + x2 - 2xz + z2] z(2x - z) (2x2 - 2xz + z2) (v) a4 - 2a2b2 + b4 (a2)2 - 2 (a2) (b2) + (b2)2 (a2 - b2)2 [(a - b) (a + b)]2 Question

Dear Student,
In (v) solution we have,a4-2a2b2+b4=a22-2a2b2+b22=a2-b22    Here, they have applied the identity x2-2xy+y2=x-y2 and in this case x=a2 and y= b2=a-ba+b2    In this step they have applied the identity x2-y2=x-yx+y=a-b2a+b2

Regards.

  • 1
What are you looking for?