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I'm very bad with math, i will go right to the question: Why i can't make this cancellation? (or why this is False, I tested this in symbolab and gave false)...

$$\frac{3^{-m}\cdot 3^{-2m+3}+27^{-m+2}}{81\cdot 3^{-m}}\:=\frac{3^{-2m+3}+27^{-m+2}}{81}$$

sorry for the sacrilege

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    $\begingroup$ You forgot to pull ($3^{-m}$) out from $27^{-m+2})$ $\endgroup$
    – Inazuma
    Apr 1, 2016 at 4:42

2 Answers 2

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What you did was akin to saying that $\dfrac{ab+ac}{ad}=\dfrac{b+ac}{d}$

The correct result should have been more along the lines of this:

$\dfrac{ab+ac}{ad}=\dfrac{a(b+c)}{ad} = \dfrac{a(b+c)}{ad}\cdot \dfrac{1/a}{1/a} = \dfrac{a(b+c)/a}{ad/a} = \dfrac{b+c}{d}$

In other words, if you are going to cancel something from numerator and denominator, you must cancel it from all terms which are being added simultaneously, not having forgotten any.

$\dfrac{3^{-m}3^{-2m+3}+27^{-m+2}}{81\cdot 3^{-m}} = \dfrac{3^{-m}3^{-2m+3}+27^{-m+2}}{81\cdot 3^{-m}}\cdot \dfrac{3^m}{3^m} = \dfrac{(3^{-m}3^{-2m+3}+27^{-m+2})3^m}{81\cdot 3^{-m}\cdot 3^m}$

$=\dfrac{3^{-2m+3}+3^m\cdot 27^{-m+2}}{81}$ which can continue to be simplified further if so desired

(continue by noting $81=3^4$ and $27=3^3$ so $27^{-m+2}=(3^3)^{-m+2}=3^{-3m+6}$...)

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From your approach:

$\frac{3^{-m}\cdot 3^{-2m+3}+27^{-m+2}}{81\cdot 3^{-m}}\:=\frac{3^{-2m+3}+3^{6-2m}}{81}$

Since $\frac{27^{-m+2}}{3^{-m}} = 3^{6-2m}$

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