Question 885807
1)
{{{(x+1)*(x-1)=x*(x-1)+1*(x-1)=x^2-x+x-1=x^2-1}}}
{{{(x+1)*(x^2-x+1)=x*(x^2-x+1)+1*(x^2-x+1)=x^3-x^2+x+x^2-x+1=x^3+1}}}
{{{(x+1)*(x^3-x^2+x-1)=x*(x^3-x^2+x-1)+1*(x^3-x^2+x-1)=x^4-x^3+x^2-x+x^3-x^2+x-1=x^4-1}}}
{{{(x+1)*(x^4-x^3+x^2-x+1)=x*(x^4-x^3+x^2-x+1)+1*(x^4-x^3+x^2-x+1)}}}=?
Those two products would be too long to write in one line, so I will add them up vertically:
{{{drawing(150,100,0,4.5,-3,0,
locate(0,0,x^5-x^4+x^3-x^2+x),
locate(0.55,-0.9,"+"),
locate(0.85,-0.7,x^4-x^3+x^2-x+1),
line(0,-1.6,4.5,-1.6),
locate(0,-1.7,x^5),locate(3.6,-2,"+"),locate(4,-2,1)
)}}}
So, {{{(x+1)*(x^4-x^3+x^2-x+1)=X^6+1}}}
 
2) In all the products above {{{(x+1)}}} is one factor,
and the other factor is a polynomial with sign that alternate,
so all the middle products cancel out.
You are left with just the product of the first terms plus the product of the last terms.
So, {{{(x+1)*(x^8-x^7+x^6-x^5+x^4-x^3+x^2-x+1)=x^9+1}}}
All the other products cancel out and you are left with the product of the {{{x}}} in {{{(x+1)}}} times the {{{x^8}}} ,
plus the product of the {{{"+1"}}} and {{{"+1"}}} at the end of both factors.
 
3) I believe what you meant to write was
{{{(x+1)*(x^100-x^99+x^98-x^97+ "..." +x^2-x+1)}}} , because if the signs alternate,
all the terms with even exponents will follow a {{{"+"}}} sign.
{{{(x+1)*(x^100-x^99+x^98-x^97+ "..." +x^2-x+1)=x^101+1}}}