[" 47.Simplify: "[2^((1)/(ab))]^(c)[2^((1)/(bc))]^(a)[2^((1)/(ac))]^(b)" when "a^(2)+b^(2)+],[c^(2)=2abc]

(a+b+c)(a^(2)+b^(2)+c^(2)-ab-bc-ac)

((1)/(a)-(1)/(b+c))/((1)/(a)+(1)/(b+c))(1+(b^(2)+c^(2)-a^(2))/(2bc)):(a-b-c)/(abc)

If (1)/(a)+(1)/(b)+(1)/(c)=1 and abc=2 then find ab^(2)c^(2)+a^(2)bc^(2)+a^(2)b^(2)c is

If A=[(a^(2),ab,ac),(ab,b^(2),bc),(ac,bc,c^(2))] and a^(2)+b^(2)+c^(2)=1 , then A^(2)=

If a,b, and c are in H.P.then th value of ((ac+ab-bc)(ab+bc-ac))/((abc)^(2)) is ((a+c)(3a-c))/(4a^(2)c^(2)) b.(2)/(bc)-(1)/(b^(2)) c.(2)/(bc)-(1)/(a^(2)) d.((a-c)(3a+c))/(4a^(2)c^(2))

|(1,a^(2)+bc,a^(3)),(1,b^(2)+ac,b^(3)),(1,c^(2)+ab,c^(3))|=-(a-b)(b-c)(c-a)(a^(2)+b^(2)+c^(2))

If (a^(2)-bc)/(a^(2) +bc) + (b^(2)-ac)/(b^(2) + ac) + (c^(2)-ab)/(c^(2)+ab)= 1 then find (a^(2))/(a^(2) + bc) + (b^(2))/(b^(2) + ac) + (c^(2))/(c^(2) +ab)= ?