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 a^(4) + b^(4) = a^(2) b^(2) find a^(6) + b^(6)

If c^(2) = a^(2) + b^(2) , then prove that 4s (s - a) (s - b) (s - c) = a^(2) b^(2)

Let R_(1) and R_(2) be two relation defined as follows : R_(1) = { (a,b) in R^(2) , a^(2) + b^(2) in Q} and R_(2) = {(a,b) in R^(2) , a^(2) + b^(2) cancel(in)Q) where Q is the set of the rational numbers. Then:

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

Find the value of ( a^2/b^2 + b^2/a^2 ) is ( a ) ( a/b + b/a )^2 - 2 ( b ) ( a/b + b/a )^2 + 2 ( c ) ( a/b + b/a )^2 + 4 ( d ) ( a/b + b/a )^2 - 4

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 :

If 4(a+2)^(2)-9a+2(2a^(2)-1)=4(b+2)^(2)-9b+2(2b^(2)-1)=15 where a,b in R and a>b then find the value of 8(a-b)

If c^(2)=a^(2)+b^(2),2s=a+b+c, then 4s(s-a)(s-b)(s-c)

Solve the following pair of linear equations: (i) p x+q y=p q ;" "q x p y=p+q (ii) a x+b y=c ;" "b x+a y=1+c (iii) x/a-y/b=0 ; a x+b y=a^2+b^2 (iv) (a-b)x+(a+b)y=a^2-2a b-b^2; (a+b)(x+y)=a^2+b^2 (v) 152 x 378 y= 74 ;" " 378

If |0a b^2a c^2a^2b0b c^2a^2cc b^2 0|=2a^pb^q c^r ,t h e np+q+r+10 is equal to 27 (b) 18 (c) 19 (d) 8