How do you divide #(x^6-y^6)/(x-y)#?
2 Answers
The following identities are true for every real
Hence
from the last equation we have that
Explanation:
Both the numerator and denominator are homogeneous polynomials in
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^n-y^n)/(x-y) = sum_(k=0)^(n-1) x^(n-1-k) y^k#