Question

How do you divide `(x^6-y^6)/(x-y)`?

Answer 1

The following identities are true for every real `a,b`

`a^2 - b^2 = (a + b)(a - b)`
`a^3 - b^3 = (a - b)(a^2 + ab + b^2)`
`a^3 + b^3 = (a + b)(a^2 - ab + b^2)`

Hence
#x^6 - y^6= = (x^3)^2 - (y^3)^2 = (x^3 + y^3)(x^3 - y^3) = (x + y)(x^2 - xy + y^2)(x - y)(x^2 + xy + y^2)#

from the last equation we have that

#x^6 - y^6= (x + y)(x^2 - xy + y^2)(x - y)(x^2 + xy + y^2)=> (x^6-y^6)/(x-y)=(x+y)*(x^2-xy+y^2)(x^2+xy+y^2)#

Answer 2

`(x^6-y^6)/(x-y) = x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5`

Explanation:

Both the numerator and denominator are homogeneous polynomials in `x` and `y`; the numerator being of degree `6` and the denominator of degree `1`.

The numerator is in standard form, but most of the possible coefficients are zero. In full, it could be written as:

`x^6+0x^5y+0x^4y^2+0x^3y^3+0x^2y^4+0xy^5-y^6`

We can long divide the coefficients like this:

which tells us that the quotient is exact and equal to:

`x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5`

In fact, in the general case we find:

`(x-y) sum_(k=0)^(n-1) x^(n-1-k) y^k`

`=x sum_(k=0)^(n-1) x^(n-1-k) y^k - y sum_(k=0)^(n-1) x^(n-1-k) y^k`

`=sum_(k=0)^(n-1) x^(n-k) y^k - sum_(k=0)^(n-1) x^(n-1-k) y^(k+1)`

`=sum_(k=0)^(n-1) x^(n-k) y^k - sum_(k=1)^n x^(n-k) y^k`

`=x^n + color(red)(cancel(color(black)(sum_(k=1)^(n-1) x^(n-k) y^k))) - color(red)(cancel(color(black)(sum_(k=1)^(n-1) x^(n-k) y^k))) - y^n`

`=x^n-y^n`

So dividing both ends by `(x-y)`, we find:

`(x^n-y^n)/(x-y) = sum_(k=0)^(n-1) x^(n-1-k) y^k`