Calculate sum of all factors of expression

Question

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..

Answer 1

$$\require{cancel} \left(\dfrac{\cancel{2}x}{\cancel{2}}\right)^2\cdot\left(\frac {\cancel{3}x}{\cancel{3}}\right)^3 = x^2\cdot x^3 = x^{2+3} = x^5 = \underbrace{x\cdot x\cdot x\cdot x\cdot x}_{\large 5\,\text{factors, each }x}$$

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

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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.