Fiery "Cubes"

$\displaystyle 2000^3-1999 \cdot 2000^2 - 1999^2 \cdot 2000 + 1999^3$

$=2000^2(2000-1999)-1999^2(2000-1999)$

$=2000^2-1999^2$

$=(2000+1999)(2000-1999)$

$=3999$

y = 2000 , x = 1999

The given expression = y^3 - x y^2 - y x^2 + x^3

= (y^3 - x y^2) - (y x^2 - x^3)

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

= (1)(1)(3999) = 3999

I said that x = 2000, then the equation becomes: ${ =\quad x }^{ 3 }-(x-1){ x }^{ 2 }-{ (x-1) }^{ 2 }x+{ (x-1) }^{ 3 }\\ =\quad { x }^{ 3 }-({ x }^{ 3 }-{ x }^{ 2 })-({ x }^{ 3 }-2{ x }^{ 2 }+x)+{ ({ x }^{ 3 }-{ 3x }^{ 2 }+3x-1) }\\ =\quad { x }^{ 3 }-{ { x }^{ 3 }+{ x }^{ 2 }- }{ x }^{ 3 }{ +2x }^{ 2 }-x+{ x }^{ 3 }-{ 3x }^{ 2 }+3x-1\\ =\quad { (x }^{ 3 }-{ { x }^{ 3 }) }+{ (x }^{ 3 }-{ { x }^{ 3 } })+({ x }^{ 2 }+2{ x }^{ 2 }-{ 3{ x }^{ 2 }) }-x+3x-1\\ =\quad 2x-1\\ =\quad 2\cdot 2000-1\\ =\quad 4000-1\\ =\quad 3999$

let 2000=x and 1999=y;

>x^3-yx^2-(y^2)x+y^3=x^2(x-y) -y^2(x-y) =(x-y)(x^2-y^2) =(x-y)(x-y)(x+y) =(2000-1999)(2000-1999)(2000+1999)=3999

200^3-1999(2000^2)-1999^2(2000)+1999^3 let 2000=x
we have;
x^3 - (x-1)(x^2) - (x-1)^2(x) + (x-1)^3 =X^3 - (x^3-x^2) - (x^3-2x^2+x) + (x^3-3x^2+3x-1) =x^3 - x^3 + x^2 - x^3 + 2x^2 - x + x^3 -3x^2 + 3x-1 =(x^3 - x^3) + (x^2 + 2x^2 - 3x^2) + (-x^3 + x^3) ( -x + 3x ) - 1 =2x-1

substitute 2000 to x 2(2000)-1 =4000-1 =3999

Let k=1000; so 2000(^3)=2k^3. Equation becomes (2k)^3-(2k-1)(2k)^2-(2k-1)^2(2k)+(2k-1)^3. Multiply using algebra and combine like terms you end up with 4k-1 = 4000 -1= 3999. remember4k means the product of 4 and 1000 or4000.