how's this?
thanks.
y^3-gy+5
y^3-3y-2
and
x^5+y^5
x+y
If you can show the solution, better.
THANKS!
You could have simply tried long division
But I am going to try a different approach...
$\displaystyle \frac{x^5 + y^5}{x+y} = y^4\frac{\left(\frac{x}{y}\right)^5 + 1}{\left(\frac{x}{y}\right)+1} = y^4\frac{-a^5 + 1}{-a+1}$
Where $\displaystyle -a = \frac{x}{y}$.
You know the geometric progression formula, $\displaystyle \frac{a^n - 1}{a - 1} = 1 + a + a^2 + a^3+ \cdots + a^{n-1}$
Thus, now we can substitute $\displaystyle a = -\frac{x}{y}$:
$\displaystyle y^4\frac{-a^5 + 1}{-a+1} = y^4(1 + a + a^2 + a^3 + a^4) = $$\displaystyle y^4(1 - \frac{x}{y} + \left(-\frac{x}{y}\right)^2 - \left(\frac{x}{y}\right)^3 + \left(-\frac{x}{y}\right)^4) = y^4 - xy^3+ x^2y^2 - x^3y + x^4 $
So:
$\displaystyle \frac{x^5 + y^5}{x+y} = y^4 - xy^3+ x^2y^2 - x^3y + x^4$
Hello, ellenbaggao!
$\displaystyle \frac{y^3-{\color{red}g}y+5}{y^3-3y-2}$ . . . . What is that $\displaystyle {\color{blue}g}$ ?
Factor: .$\displaystyle \frac{(x+y)(x^4-x^3y + x^2y^2 - xy^3 + y^4)}{x+y}$$\displaystyle \frac{x^5+y^5}{x+y}$
Reduce: .$\displaystyle x^4 - x^3y + x^2y^2 - xy^3 + y^4$