# Difference of Squares

• Sep 4th 2008, 05:20 PM
RubyRed
Difference of Squares
How do I show that $\displaystyle \frac{(x+\frac{1}{x})^6 - (x^6+\frac{1}{x^6}) -2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})}$

equals

$\displaystyle (x+\frac{1}{x})^3-(x^3+\frac{1}{x^3})$

and then how that equals

$\displaystyle 3(x+\frac{1}{x})$

Thanks!
• Sep 4th 2008, 06:40 PM
Soroban
Hello, RubyRed!

Quote:

How do I show that: .$\displaystyle \frac{\left(x+\dfrac{1}{x}\right)^6 - \left(x^6+\dfrac{1}{x^6}\right) -2}{\left(x+\dfrac{1}{x}\right)^3+\left(x^3+\dfrac{ 1}{x^3}\right)} \;\;=\;\;\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)$

and then how that equals: .$\displaystyle 3\left(x+\frac{1}{x}\right)$

The numerator is: .$\displaystyle \left(x + \frac{1}{x}\right)^6 - \left(x^6 + 2 + \frac{1}{x^6}\right) \;=\;\left(x + \frac{1}{x}\right)^6 - \left(x^3 + \frac{1}{x^3}\right)^2$

. . . . $\displaystyle = \;\bigg[\left(x + \frac{1}{x}\right)^3\bigg]^2 - \left[x^3 + \frac{1}{x^3}\right]^2$ . . . a difference of squares

which factors: .$\displaystyle \bigg[\left(x + \frac{1}{x^3}\right)^3 - \left(x^3 + \frac{1}{x^3}\right)\bigg]\cdot \bigg[\left(x + \frac{1}{x}\right)^3 + \left(x^3 + \frac{1}{x^3}\right)\bigg]$

The fraction becomes: .$\displaystyle \frac{ \bigg[\left(x + \frac{1}{x^3}\right)^3 - \left(x^3 + \frac{1}{x^3}\right)\bigg]\cdot \bigg[\left(x + \frac{1}{x}\right)^3 + \left(x^3 + \frac{1}{x^3}\right)\bigg] } {\bigg[\left(x+\frac{1}{x}\right)^3 + \left(x^3 + \frac{1}{x^3}\right)\bigg]}$

. . which reduces to: .$\displaystyle \boxed{\left(x+\frac{1}{x}\right)^3 - \left(x^3 + \frac{1}{x^3}\right)}$

Expand the cube: .$\displaystyle \left(x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}\right) - \left(x^3 + \frac{1}{x^3}\right)$

. . . $\displaystyle =\;\;x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} - x^3 - \frac{1}{x^3} \;\;=\;\;3x + \frac{3}{x} \;\;=\;\;\boxed{3\left(x + \frac{1}{x}\right)}$