(a+b+c)^(2)=

(cos ^(2) ((B-C)/( 2)) )/( (b+c)^(2))+( sin ^(2)((B-C)/( 2)) )/( (b-c)^(2))=

(cos^(2)((B-C)/(2)))/((b+c)^(2))+(sin^(2)((B-C)/(2)))/((b-c)^(2))=(1)/(a^(2))

Simplify: (a^(2)-(b-c)^(2))/((a+c)^(2)-b^(2))+(b^(2)-(a-c)^(2))/((a+b)^(2)-c^(2))+(c^(2)-(a-b)^(2))/((b+c)^(2)-a^(2))

Prove that |{:((b+c)^(2), a^(2), bc),((c+a)^(2), b^(2), ca),((a+b)^(2), c^(2), ab):}|= (a-b) (b-c)(c-a)(a + b+c) (a^(2) + b^(2) + c^(2)) .

If a,b,c gt 0, then (a )/(2a + b +c ) = (b)/(a + 2b +c)= (c )/(a +b + 2c) = ?

IF a+b+c=0 then (a^2+b^2+c^2)/((a-b)^2+(b-c)^2+(c-a)^2) =

Factorise : (a^(2)-b^(2))(a+b)+(b^(2)-c^(2))(b+c)+(c^(2)-a^(2))(c+a)

prove : ((2^(a))/(2^(b)))^(a+b)*((2^(b))/(2^(c)))^(b+c)*((2^(c))/(2^(a)))^(c+a)=1