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Expression: $$ \left(\frac{2x}2\right)^2 \left(\frac{3y}3\right)^3$$

Sum of all factors of above expression is $$2\cdot \left(\dfrac{2x}2\right) + 3\cdot\left(\dfrac {3y}3\right)$$

How ?

Can anyone one please explain how do we arrive at this..

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    $\begingroup$ The sum of factors of $a^2b^3$ is $a+a+b+b+b$. $\endgroup$ – Yves Daoust Nov 2 '14 at 14:55
  • $\begingroup$ @YvesDaoust: $a^2b^3$ has $3\times4=12$ factors. Namely $1,a,b,a^2,b^2,b^3,ab,ab^2,ab^3,a^2b,a^2b^2,a^2b^3$ $\endgroup$ – Bumblebee Nov 2 '14 at 14:58
  • $\begingroup$ Where did the $y$ come from in the second line? Was the second part of the first supposed to be $(3y/3)^3?$ Why do you not write this as $y^3$? $\endgroup$ – Ross Millikan Nov 2 '14 at 15:01
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    $\begingroup$ @Nilan: $3x/3 = \bf x$: typo to fix so as not to cause confustion. $\endgroup$ – amWhy Nov 2 '14 at 15:05
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    $\begingroup$ @Nilan: you are listing all divisors, not all of which are single factors. My comment is in line with the answer given by the OP, if you expand the expression as a product without powers, but keeping the parenthesis ($aabbb$). $\endgroup$ – Yves Daoust Nov 2 '14 at 15:17
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$$\require{cancel} \left(\dfrac{\cancel{2}x}{\cancel{2}}\right)^2\cdot\left(\frac {\cancel{3}x}{\cancel{3}}\right)^3 = x^2\cdot x^3 = x^{2+3} = x^5 = \underbrace{x\cdot x\cdot x\cdot x\cdot x}_{\large 5\,\text{factors, each }x}$$

Summing the five factors of $x$ gives us $x + x+x+x+x = 5x$


Edited answer since the post has been edited

Now we have $$\left(\dfrac{\cancel{2}x}{\cancel{2}}\right)^2\cdot\left(\frac {\cancel{3}y}{\cancel{3}}\right)^3 = x^2\cdot y^3= x \cdot x \cdot y \cdot y \cdot y $$

$$\text{Sum of factors }\,x + x + y + y+y = 2x + 3y$$

*Note that this second answer matches the answer in your post:

$$ \left(\frac{2x}{2}\right)^2 + \left(\frac{3y}3\right)^3 = \frac{2x}{2}\cdot \frac{2x}{2}\cdot \frac{3y}3\cdot \frac{3y}3\cdot \frac{3y}3$$

Now summing the five factors: $$\frac{2x}{2}+ \frac{2x}{2}+ \frac{3y}3+ \frac{3y}3+\frac{3y}3 = \cancel{2}\left(\frac{2x}{\cancel{2}}\right) + \cancel{3}\left(\frac{3y}{\cancel{3}}\right)= 2x + 3y$$

See how canceling the common factors in the numerator and denominator simplified matters.

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