Simplifying Expression with Positive Variables a, b, and c | Step-by-Step Guide

  • Thread starter Thread starter BruceSpringste
  • Start date Start date
AI Thread Summary
The discussion focuses on simplifying the expression x = ((ab√c)^(1/3) - a(b^2c)^(1/4)) / (a^(3)b^(2)c)^(1/6) given that a, b, and c are positive variables. The initial confusion is addressed by substituting √c with c^(1/2) and applying laws of exponents to break down the expression. The simplification process involves factoring out the largest powers of a, b, and c from the numerator and canceling with the denominator. The final expression is presented in a clearer format using LaTeX, demonstrating that the simplification is more manageable than initially perceived. Overall, the thread highlights the importance of understanding exponent rules in simplifying complex expressions.
BruceSpringste
Messages
37
Reaction score
0

Homework Statement


Om a > 0, b > 0, c > 0, och x = ((ab√c)1/3-a(b2c)1/4)/(a3b2c)1/6

The Attempt at a Solution



I have no idea where to start. I understand the relevance of a,b and c being > 0 in order to simplify but other than that I am pretty much stuck!
 
Physics news on Phys.org
A first obvious step is to replace that \sqrt{c} with c^{1/2}. Then use the "laws of exponentials: (abc^{1/2})^{1/3}= a^{1/3}b^{1/3}c^{1/6}, a(b^2c)^{1/4}= ab^{1/2}c^{1/4} and (a^3b^2c)^{1/6}= a^{1/2}b^{1/3}c^{1/6}

So you have \frac{a^{1/3}b^{1/3}c^{1/6}- ab^{1/2}c^{1/4}}{a^{1/2}b^{1/3}c^{1/6}}

Factor the largest power of a, b, and c in both terms in the numerator and cancel what you can with the denominator.
 
  • Like
Likes 1 person
\begin{align*}
\frac{\sqrt[3]{ab \sqrt{c}} - a \sqrt[4]{b^2 c}}{\sqrt[6]{a^3 b^2 c}}
\end{align*}

Wrote the equation in latex too, it was easier to see.
 
Alright that was easier than I previously thought! Thanks!
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top