2 Equalities in Different Veins
by agbdmrbirdyface, Aug 25, 2016, 12:37 AM
Both of these problems were problems I solved way back in spring for a math camp during spring break (we don't question why).
Solution:
Solution:
Areteem Institute, Algebraic Transformations wrote:
Given that
, show that



Solution:
This, on first sight, could be a very, very bad bash. 
Perhaps we can find a very nice way to exploit the
condition. On that note, let's multiply all of these fractions by
!
Let's multiply the first fraction by
, the second one by
, the next one by
, and the last one by
.
This will give us:

And we can apply the
condition:


which is obvious.

Perhaps we can find a very nice way to exploit the


Let's multiply the first fraction by




This will give us:

And we can apply the



which is obvious.
Areteem Institute, Algebraic Transformations wrote:
Given that
and
, prove that ![$\sqrt[3]{1991} + \sqrt[3]{1992} + \sqrt[3]{1993} = \sqrt[3]{1991x^2 + 1992y^2 + 1993z^2}$](//latex.artofproblemsolving.com/c/3/a/c3a9662533afc15ac552514f808e6c35a84c7715.png)


![$\sqrt[3]{1991} + \sqrt[3]{1992} + \sqrt[3]{1993} = \sqrt[3]{1991x^2 + 1992y^2 + 1993z^2}$](http://latex.artofproblemsolving.com/c/3/a/c3a9662533afc15ac552514f808e6c35a84c7715.png)
Solution:
Again, tempted to do another bad bash by cubing.
.
Maybe we could use the equality given to our advantage. Suppose we take the cube root of this equality and set it equal to some constant
.
, and also 
Now we can write both sides of the equality in
.
![$\frac{k}{x} + \frac{k}{y} + \frac{k}{z} = \sqrt[3]{\frac{k^3}{x} + \frac{k^3}{y} + \frac{k^3}{z}}$](//latex.artofproblemsolving.com/c/7/4/c745b865134d00791f43dc74d0c36968f1f8e0ce.png)
![$\Rightarrow k(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = \sqrt[3]{k^3 (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})}$](//latex.artofproblemsolving.com/9/2/9/9298dc5a254345f8f7ebf5324628a4b2c4ffd905.png)
![$\Rightarrow k = \sqrt[3]{k^3} = k$](//latex.artofproblemsolving.com/0/b/1/0b1609e4ba35e6bfd02f4bdb2014b8eb819296e6.png)
Simplifying the entire thing finishes the proof.

Maybe we could use the equality given to our advantage. Suppose we take the cube root of this equality and set it equal to some constant

![$\sqrt[3]{1991} x = \sqrt[3]{1992} y = \sqrt[3]{1993} z = k$](http://latex.artofproblemsolving.com/f/7/c/f7c72826335ee17fa6760351774542291aea5405.png)

Now we can write both sides of the equality in

![$\frac{k}{x} + \frac{k}{y} + \frac{k}{z} = \sqrt[3]{\frac{k^3}{x} + \frac{k^3}{y} + \frac{k^3}{z}}$](http://latex.artofproblemsolving.com/c/7/4/c745b865134d00791f43dc74d0c36968f1f8e0ce.png)
![$\Rightarrow k(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = \sqrt[3]{k^3 (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})}$](http://latex.artofproblemsolving.com/9/2/9/9298dc5a254345f8f7ebf5324628a4b2c4ffd905.png)
![$\Rightarrow k = \sqrt[3]{k^3} = k$](http://latex.artofproblemsolving.com/0/b/1/0b1609e4ba35e6bfd02f4bdb2014b8eb819296e6.png)
Simplifying the entire thing finishes the proof.