Factorise:1 − (b − c)2
We have:
$1-(b-c)^{2}=(1)^{2}-(b-c)^{2}$
$=\{1+(b-c)\}\{1-(b-c)\}$
$=(1+b-c)(1-b+c)$
$\therefore 1-(b-c)^{2}=(1+b-c)(1-b+c)$
Leave a comment
All Study Material
