# Calculate (x+y)^2 + (x+y+1)^2 - (3x^2+4xy) + (y-1) =... x^2003-1 - (x -1)(x^2002+ ...+ x+1) =...

### 2 Answers | Add Yours

1)

(x+y)^2 + (x+y+1)^2 - (3x^2+4xy) + (y-1) =

= (x+y)^2+(x+y+1)^2 - (3x^2+4xy)+(y-1)

= (x+y)^2 + (x+y)^2 +2(x+y)+1 - 3x^2 -4xy +(y-1)

=2x^2+4xy +2y^2 +2x+2y+1-3x^2 -4xy +y-1

= -x^2 +2y^2 +2x+3y

2)

To calculate x^2003-1 - (x -1)(x^2002+ ...+ x+1) = ...

We know that (x-1){x^2002+x^2001+... x^2+x+1) =

(x^2003+x^2002+x^2001+.....x^3+x^2+x)- (x^2002+x^2001+x^2000+.....+x^3+x^2+x+1)

= x^2003-1.

Therefore the given expression= x^2003-1) - (x-1){x^2002+x^2001+x^2002+....x^2+x+1} = (x^2003-1)-(x^2003-1) = 0.

1) To calculate the first expression, we'll expand the squares, using the formulas:

(a+b)^2 = a^2 + 2ab + b^2

(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+ac+bc)

(x+y)^2 = x^2 + 2xy + y^2

(x+y+1)^2 = x^2 + y^2 + 1 + 2(xy + x + y)

We'll re-write the expressions substituting the squares:

E (x,y) = x^2 + 2xy + y^2 + x^2 + y^2 + 1 + 2(xy + x + y) - (3x^2 + 4xy) + (y-1)

Now, we'll remove the brackets and combine like terms:

E(x,y) = 2x^2 + 2y^2 + 2xy + 1 + 2xy + 2x + 2y - 3x^2 - 4xy + y - 1

We'll combine and eliminate like terms:

E(x,y) = 2x^2 + 2y^2+ 2x + 2y - 3x^2 + y

**E(x,y) = -x^2 + 2y^2 + 2x + 3y**

2) Now, we'll calculate the expression:

x^2003 - 1 - (x -1)(x^2002+ ...+ x+1) =...

We'll use the formula:

x^n - 1 = (x-1)(x^(n-1) + x^(n-2) + ......... + x + 1)

We'll substitute n by 2003

x^2003 - 1 = (x-1)(x^2002 + x^2001 + ........ + x + 1)

We'll re-write the expression:

E (x) = (x-1)(x^2002 + x^2001 + ........ + x + 1) - (x -1)(x^2002+ ...+ x+1)

We'll eliminate like terms: