Question

factorise the expression a^2-b^2-c^2-2bc+a+b+c ​

Answer 1

Step-by-step explanation:

a2+b2+c2+2ab−2ac−2bc=−(b+c)2

⟹(a)2+(b)2+(−c)2+2(a)(b)+2(a)(−c)+2(b)(−c)+(b+c)2=0

⟹(a+b−c)2+(b+c)2=0

⟹(a+b−c)2−{(b+c)i}2=0

⟹{a+b−c−(b+c)i}{a+b−c+(b+c)i}=0

⟹{a+b(1−i)−c(1+i)}{a+b(1+i)−c(1−i)}=0

Answer 2

Use Identity :-

$= > {(a + b + c)}^{2}$

Then...

$= > (a + b + c) \times (a + b + c)$

$= > a(a + b + c) + b(a + b + c) + c(a + b + c)$

$= > ({a}^{2} + ab + ac) + (ab + {b}^{2} + bc) + (ac + bc + {c}^{2})$

$= > {a}^{2} + ab + ac + ab + {b}^{2} + bc + ac + bc + {c}^{2}$

$= > {a}^{2} + 2ab + 2ac + {b}^{2} + 2bc + {c}^{2}$

$= > {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ac$

Hence, Proved...☺️