# Thread: How are these equal? Last question.

1. ## How are these equal? Last question.

Show that $\displaystyle (2^\frac{1}{3})^{-2} +2^\frac{1}{3} -2 = \frac{3*2^\frac{1}{3}}{2} - 2$.

I usually use my calculator to convert from one form to another, but I don't actually see or understand how it is really the same. Any help would be appreciated.

2. Hi, I will show you one way of doing this.

I multiply the whole thing by $\displaystyle 2$,

$\displaystyle 2^{-2/3}\cdot 2^1 + 2^{1/3}\cdot 2^1 - 2^2 = 3\cdot 2^{1/3} - 2^2$

Then by using the fact that 2$\displaystyle 2^a\cdot 2^b = 2^{a+b}$ we get,

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

Finally we get,

$\displaystyle 2^{4/3} = 2^{1/3}(3-1)$,

and I hope the rest is clear.

3. Another way to do it:
$\displaystyle (2^\frac{1}{3})^{-2} +2^\frac{1}{3}= 2^{-2/3}+ 2^{1/3}= \left(2^{1/3}\right)\left(2^{-3/3}+ 1\right)$$\displaystyle = \left(2^{1/3}\right)\left(\frac{1}{2}+ 1\right)= \left(2^{1/3}\right)\left(\frac{3}{2}\right)$. Now subtract 2 from both sides.

4. Hello, Joker37!

I went at it head-on . . . no fancy tricks.

Show that: . $\displaystyle (2^\frac{1}{3})^{-2} +2^\frac{1}{3} -2 \;=\; \dfrac{3\cdot2^\frac{1}{3}}{2} - 2$

$\displaystyle (2^{\frac{1}{3}})^{-2} + 2^\frac{1}{3}} - 2 \;\;=\;\;2^{-\frac{2}{3}} + 2^{\frac{1}{3}}} - 2$

. . . . $\displaystyle \displaystyle =\;\;\dfrac{1}{2^{\frac{2}{3}}} + 2^{\frac{1}{3}} - 2 \;\;=\;\;\frac{1}{2^{\frac{2}{3}}}\cdot\frac{2^{\f rac{1}{3}}}{2^{\frac{1}{3}}} + 2^{\frac{1}{3}}\cdot\frac{2}{2} - 2$

. . . . $\displaystyle \displaystyle =\;\; \frac{2^{\frac{1}{3}}}{2} + \frac{2\cdot2^{\frac{1}{3}}}{2} - 2 \;\;=\;\;\frac{3\cdot2^{\frac{1}{3}}}{2} - 2$

5. You can also manipulate such that you get 3 / 2^(2/3) ; try it.

Curious: why in heck have "-2" on each side ?!