X^4-(X^3)y+(X^2)(y^2)-x(y^3)+y^4= (x+y)^4-5xy(x+y)^2+5(xy)^2

x^6-(x^5)y+(X^4)(y^2)-(X^3)(y^3)+(X^2)(y^4)-x(y^5)+y^6=(x+y)^6-7xy(x+y)^4 +14((xy)^2)(x+y)^2 - 7(xy)^3

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If n is an odd prime then prove,

x^n-1 - X^(n-2).y+..........-x.y^(n-2)+y^(n-1) = (x+y)^n-1 - nxy(x+y)^(n-3) +..........(-1)^((n-1)/2) . n .(xy) ^ ((n-1)/2)

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X^4-(X^3)y+(X^2)(y^2)-x(y^3)+y^4= (x+y)^4-5xy(x+y)^2+5(xy)^2

x^6-(x^5)y+(X^4)(y^2)-(X^3)(y^3)+(X^2)(y^4)-x(y^5)+y^6=(x+y)^6-7xy(x+y)^4 +14((xy)^2)(x+y)^2 - 7(xy)^3

....................
If n is an odd prime then prove,

x^n-1 - X^(n-2).y+..........-x.y^(n-2)+y^(n-1) = (x+y)^n-1 - nxy(x+y)^(n-3) +..........(-1)^((n-1)/2) . n .(xy) ^ ((n-1)/2)

Your post is very difficult to read in ASCII...I even didn't try. My advice: learn how to post here in LaTeX or else

attach some document, preferably PDF, where mathematical stuff appears decently.

DonAntonio

I think this has been posted before recently but, for the love of me, I can't find the original post.

And I believe the suggestion was to learn about the binomial theorem; then try to expand (x+y)^4 and (x+y)^6.

that's simple re arrangement of binomial expansions. for general proof, try general form of binomial expansions. you will get it easily.

Or else use the "sup" and "sub" tags to create superscripts and subscripts:

[noparse]xy xy[/noparse]

xy xy

Got the point. Thanks every one...