HCF and LCM
H.C.F
Q. Find H.C.F of: a4+a2b2+b4, a3-b3, and a3+a2b +ab2
Soln:
1St exp: a4+a2b2+b4
= (a2)2 + ( b2 )2 + a2b2
= (a2 + b2 )2 - 2a2b2 + a2 b2
= (a2 + b2)2 - ab2
= (a2+ ab +b2) (a2- ab + b2)
2Nd exp: a3 - b 3
= a - b (a2+ ab + b2)
3rd exp: a3+ a2b + ab2
= a (a2+ ab + b2 )
Therefore, H.C.F = (a2+ ab + b2 )