If `P=a^2-b^2+2a b ,\ Q=a^2+4b^2-6a b\ ,` `R=b^2+b ,\ S=a^2-4a b` and `T=-2a^2+b^2-a b+adot` Find `P+Q+R+S-Tdot`

If relation R_1={(a,b):a,b in R,a^2+b^2 in Q} and R_2={(a,b):a,b in R, a^2+b^2 !in Q} Then

To find the square of the algebric expressions given below using (a-b)^2 = a^2- 2ab + b^2 , what has to be susbtitued for a and b in each cases and hence let's find the squares. p + q - r - s

If a(p+q)^2+2b p q+c=0abd a(p+r)^2+2b p r+c=0(a!=0) , then a. q r=p^2 b. q r=p^2+c/a c. q r=p^2 d. none of these

For every a, b in N a" @ "b=a^(2)" when "(a+b)" is even" =a^(2)-b^(2)" when "(a+b)" is odd" a#b=b^(2)" when "axxb" is odd" =a^(2)-b^(2)" when "axxb" is even" Find the value of (p#q)-(q" @ "p)-(q" @ "p),p is even and q is odd :

Let A=[{:(a,b,c),(p,q,r),(1,1,1):}]and B=A^(2) If (a-b)^(2) +(p-q)^(2) =25, (b-c) ^(2)+ (q-r)^(2)= 36 and (c-a)^(2) +(r-p)^(2)=49, then det B is

p @ q = p^2 - 20 if p > q , q 2 q = p^2 + 20 if p le q, then for a = 5 and b = 4 the value of [(a @ b) @ (a @ b)] - [(b @ a) @ (b @ a)] is :

If p=a+bomega+comega^2 , q=b+comega+aomega^2 , and r=c+aomega+bomega^2 , where a ,b ,c!=0 and omega is the complex cube root of unity, then (a) p+q+r=a+b+c (b) p^2+q^2+r^2=a^2+b^2+c^2 (c) p^2+q^2+r^2=-2(p q+q r+r p) (d) none of these

If a(p+q)^2+2b p q+c=0 and a(p+r)^2+2b p r+c=0(a!=0) , then