If `P=7x^2+5xy-9y^2 ,\ Q=4y^2-3x^2-6x y\ ` and `R=-4x^2+x y+5y^2,\ s how\ t h a t\ P+Q+R=0.`

If P=7x^2+5x y-9y^2,\ Q=4y^2-3x^2-6x y and R=-4x^2+x y+5y^2, show that P+Q+R=0.

If A = 7x^(2) + 5xy - 9y^(2), B = - 4x^(2) + xy + 5y^(2) and C = 4y^(2) - 3x^(2) - 6xy then show that A + B + C = 0

Subtract: x^2-3x y+7y^2-2 from 6x y-4x^2-y^2+5

If P= (x^3+y^3)/((x-y)^2+3xy), Q= ((x+y)^2-3xy)/(x^3-y^3) and R=((x+y)^2+(x-y)^2)/(x^2-y^2) , then (P div Q) xx R is equal to: यदि P= (x^3+y^3)/((x-y)^2+3xy), Q= ((x+y)^2-3xy)/(x^3-y^3) तथा R=((x+y)^2+(x-y)^2)/(x^2-y^2) है, तो (P div Q) xx R का मान क्या होगा ?

Let P = 12xy -10y^(2) - 18x^(2) Q = 14x^(2) + 12y^(2) + 9xy, and R = 5y^(2) -x^(2) + xy then (P + Q) - R =

If P=3x-4y-8z, Q=-10y+7x+11z and R=19z-6y+4x , then P-Q+R is equal to

If P_(1), P_(2), P_(3) are the perimeters of the three circles. S_(1) : x^(2) + y^(2) + 8x - 6y = 0 S_(2) , 4x^(2) + 4y^(2) -4x - 12y - 186 = 0 and S_(3) : x^(2) + y^(2) -6x + 6y - 9 = 0 repeectively, then the relation amongst P_(1), P_(2) and P_(3) is .............

In the algebraic expression 5x^2y+7x y^2-3x y-4y x^2, we have 5x^2y and -4y x^2 as like terms, whereas 7x y^2 and -3x y are unlike terms.

Add: (i) x y^2+4x^2y-7x^2y-3x y^2+3\ a n d\ x^(2y)+x y^2 (ii) 5x^2+7y-6z^2,\ 4y+3x^2,9x^2+2z^2-9y\ a n d\ 2y-2x^2