Question

The value of (x+3y)whole square +(x-3y)whole square is

Answer 1

x^2 + 9y^2 + 6xy +(x^2 + 9y^2 - 6xy)

= x^2 + 9y^2 + 6xy + x^2 + 9y^2 - 6xy

= 2x^2 + 18y^2

= x^2 + 9y^2

Answer 2

Step-by-step explanation:

(a+b)whole square ie. (a+b)2=

(a)square (a)2+(b)square (b)2+2×a×b-2(ab)

(a-b)whole square ie. (a-b)2=

b)2=(a)square (a)2+(b)square (b)2-2×a×b-2(ab)

Hence

(x+3y) whole2 = x2+9y2+6xy

(x-3y) whole2= x2+9y2-6xy

Therefore (x+3y)whole2+(x-3y)whole2= 2(x)square+18(y)square

ie. (x+3y)2+(x-3y)2= 2(x)2+18(y)2.