Spain | A Nice Algebra Problem

[@عبدالواسع-س8م]

Thanks so much.

[@tigerinthejungle_14]

Easier method:-
If we take the sq.rt.27 in the numerator and make it 3*sq.rt.3 and make the sq.rt.27 in the denominator into (sq.rt.3)^3 and then we get num. and den. equal, so we can directly cancel it

[@juergenilse3259]

We can rewrite enumerator and denominator sepaately as powers of sqrt(3):
sqrt(3)^sqrt(27)=sqrt(3)^sqrt(3^3)=sqrt(3)^(3*sqrt(3))
sqrt(27)^sqrt(3)=sqrt(3^3)^sqrt(3)=(sqrt(3)^3)^sqrt(3)=sqrt(3)^(3*sqrt(3))
So enuerator and denoinator have the sae vaue, and therefor the resut is 1.

[@jax2907]

It's much easier to think of the non-exponent roots as a^(0.5).
Changes it to:
=3^(0.5*rt(27)) / 27^(0.5*rt(3))
=3^(0.5*3*rt(3)) / (3^3)^(0.5*rt(3))
=3^(1.5*rt(3) / 3^(3*0.5*rt(3))
=3^(1.5*rt(3) / 3^(1.5*rt(3)
=1

[@michaeldoerr5810]

The answer is 1. Also I have figured it out without using your method.

[@baselinesweb]

Nice problem, but can the music. Nobody is interested in that - this is a math tutorial.

[@tigerinthejungle_14]

What you did is a very long method, I got it in a second 😒😒

[@eugenstar81]

This is beauty of math. Step by step. Dance of numbers

[@tigerinthejungle_14]

@@eugenstar81 What I say is there is simple method to do it so, why should we elongate it so much

[@baselinesweb]

Whoopdie doo for you