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

2 Answers

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)#

Mar 19, 2016

#(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:

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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#