Question

The H.C.F. of m²-n² and (m-n)² is ...

Answer 1

(m-n)

Explanation:

1st expression=

`= {m}^(2) - {n}^(2) \: \: \: \: \: \: \: \: \: \:`

=(m-n) (m+n)

2nd expression=

`({m - n})^(2)`

=(m-n)(m-n)

therefore H.C.F=(m-n)

taking common (m-n)