1. ## Indices

A) if $\displaystyle x=2^{1/3} + 4^{1/3}$ show that $\displaystyle x^3 = 6(1+x)$

B) If $\displaystyle x= \frac {1}{2} + \frac {1}{2} \sqrt{5}$, show that $\displaystyle \frac {x^2+x^{-2}}{x-x^{-1}} = 3$

2. Originally Posted by deltaxray
A) if $\displaystyle x=2^{1/3} + 4^{1/3}$ show that $\displaystyle x^3 = 6(1+x)$
B) If $\displaystyle x= \frac {1}{2} + \frac {1}{2} \sqrt{5}$, show that $\displaystyle \frac {x^2+x^{-2}}{x-x^{-1}} = 3$
For part B) I would note that
$\displaystyle \frac {x^2+x^{-2}}{x-x^{-1}} = \frac {x^4+1}{x^3-x}$

3. Originally Posted by deltaxray

A) if $\displaystyle x=2^{1/3} + 4^{1/3}$ show that $\displaystyle x^3 = 6(1+x)$

B) If $\displaystyle x= \frac {1}{2} + \frac {1}{2} \sqrt{5}$, show that $\displaystyle \frac {x^2+x^{-2}}{x-x^{-1}} = 3$

$\displaystyle x^3 = \left(2^{\frac{1}{3}} + 4^{\frac{1}{3}}\right)^3$

$\displaystyle = \left(2^{\frac{1}{3}}\right)^3 + 3\left(2^{\frac{1}{3}}\right)^2\left(4^{\frac{1}{3 }}\right) + 3\left(2^{\frac{1}{3}}\right)\left(4^{\frac{1}{3}} \right)^2 + \left(4^{\frac{1}{3}}\right)^3$

$\displaystyle = 2 + 3\cdot 2^{\frac{2}{3}} \cdot 4^{\frac{1}{3}} + 3 \cdot 2^{\frac{1}{3}}\cdot 4^{\frac{2}{3}} + 4$

$\displaystyle = 6 + 3\cdot 2^{\frac{4}{3}} + 3\cdot 2^{\frac{5}{3}}$

$\displaystyle 6(1 + x) = 6\left(1 + 2^{\frac{1}{3}} + 4^{\frac{1}{3}}\right)$

$\displaystyle = 6 + 6\cdot 2^{\frac{1}{3}} + 6\cdot 2^{\frac{2}{3}}$

$\displaystyle = 6 + 3\cdot 2 \cdot 2^{\frac{1}{3}} + 3 \cdot 2 \cdot 2^{\frac{2}{3}}$

$\displaystyle = 6 + 3\cdot 2^{\frac{4}{3}} + 3\cdot 2^{\frac{5}{3}}$

$\displaystyle = x^3$.