How to prove Var [X+Y] = Var [X]+2Cov (X, Y)+Var [Y] in covariance?

Var [X + Y ] = Var [X] + 2Cov (X，Y ) + Var [Y]

Just use the definition to prove it,

Var(X)=E(X^2)-[E(X)]^2

Var(Y)=E(Y^2)-[E(Y)]^2

Cov(X，Y)=E(XY)-E(X)E(Y)

therefore

]=e[(x+y)^2]-[e(x+y)]^2=e(x^2+2xy+y^2)-[e(x)+e(y)}^2

={e(x^2)-[e(x)]^2}+{e(y^2)-[e(y)]^2}+ 2[e(xy)-e(x)e(y)]

=Var(X)+2Cov (X，Y ) + Var (Y)

X and y are independent of each other, so E(XY)=E(X)E(Y)

So Cov(X, Y)=E(XY)-E(X)E(Y)=0.

Var [X -Y]=Var(X)+2Cov (X，-Y) + Var (-Y)=Var(X)-2Cov (X，Y) + Var (Y)=Var(X)+ Var (Y)

Var [X + Y ] = Var [X] + 2Cov (X，Y ) + Var [Y]=Var(X)+ Var (Y)

So Var [X+Y] = Var [X-Y]

I don't understand the meaning of the following. Can you write clearly? .