`x/a+y/b=a^2+b^2andx/(a^2)+y/(b^2)=a+b`

If x/a + y/b = a+b and x/a^(2) + y/b^(2) =2 , then what is x/a^(2) - y/b^(2) equal to?

(x)/(a)+(y)/(b)=a^(2)+b^(2) and (x)/(a^(2))+(y)/(b^(2))=a+b

Solve: (x)/(a)+(y)/(b)=a^(2)+b^(2) and (x)/(a^(2))+(y)/(b^(2))=a+b

(x)/(a)+(y)/(b)=a+b(x)/(a^(2))+(y)/(b^(2))=2,a!=0,b!=0

Solve the system of equations for x and y:(b^(2)x)/(a)-(a^(2)y)/(b)=ab(a+b) and b^(2)x-a^(2)y=2a^(2)b

If the origin is shifted to the point ((a b)/(a-b),0) without rotation, then the equation (a-b)(x^2+y^2)-2a b x=0 becomes (A)(a-b)(x^2+y^2)-(a+b)x y+a b x=a^2 (B)(a+b)(x^2+y^2)=2a b (C)(x^2+y^2)=(a^2+b^2) (D)(a-b)^2(x^2+y^2)=a^2b^2

If (x+i y)^(1//3)=a+i b ,x ,y ,a b in Rdot Show that x/a+y/b=4(a^2-b^2) (ii) x/a-y/b=2(a^2+b^2)

x/a-y/b=0 , a x+b y=a^2+b^2