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

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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.
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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!
 
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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.
 
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\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!
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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