Question

(a+b+c),(a-b+c),(a+b-c)​

Answer 1

Answer:

(b+c+a)(b+c−a)=λbc

⇒(b+c)

2

−a

2

=λbc

⇒b

2

+c

2

+2bc−a

2

=λbc

⇒λ=2⋅

2bc

b

2

+c

2

−a

2

+2=2cosA+2=4cos

2

2

A

Now since 0<

2

A

<

2

π

⇒0<cos

2

2

A

<1

Hence λ∈(0,4)

Answer 2

Answer:

(a+b+c),(a-b+c),(a+b-c) =( a+b+c)³