Messing up my algebra seriously i need to re-learn my algebra

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The discussion centers on the need to re-learn algebra due to difficulties with a specific problem involving simplification of a rational expression. The user attempted to simplify the expression but realized mistakes were made, particularly in handling the denominator. Key advice includes the importance of factorizing terms correctly, especially in the denominator, to avoid errors. Additionally, a reminder is given that exponent rules do not apply universally, emphasizing the need for careful application of algebraic principles. Overall, the conversation highlights common pitfalls in algebraic simplification and the necessity of a solid understanding of foundational concepts.
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messing up my algebra... seriously... i need to re-learn my algebra...

The Attempt at a Solution



ok so i have.
((3x^2)-6x)/(3((x^3)-(3x^2))^(2/3))

and i simplified it too.
((3x^2)-6x)/(3((x^(5/3))-(3x^(4/3))^(2/3))

then further simplified it too.
(3x(x-2))/(3x(x^(2/3))-(3x^(1/3)))

then got
((x-2))/((x^(2/3))-(3x^(1/3)))

i found out that i am doing it wrong but i don't know what...

Thank You ahead of time. Thank You.
 
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I've no idea what you've done form line 1 to line 2. However, (a+b)^c\neq a^c+b^c, in general.

I would start by factorising the terms in the denominator.
 


The most problematic part of solving this would be when factorising the denominator:

Generally, (ax+bx)^n = (x[a+b])^n = x^n(a+b)^n
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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