how is $(3a) \times(3b)$ equal to $9ab$ and not $3ab$?

Question

I'm unsure how $3a3b$ equals $9ab$ and not $3ab$? I get here by doing:

$(a + a + a) \cdot (b + b + b)$

which becomes

$(ab + ab + ab)$

which is

$3ab$?

The only way I can get $3a3b$ to $9ab$ is by doing $33ab$ as $3 * 3 * a * b$ which is $9ab$?

But I see that more as a "trick" not a rule moving the 3? Is this the case or not?

Answer 1

$(a+a+a)\cdot (b+b+b)=a\cdot (b+b+b)+a\cdot (b+b+b)+a\cdot (b+b+b)$

Answer 2

write $$(a+a+a)(b+b+b)=ab+ab+ab+ab+ab+ab+ab+ab+ab=9ab$$

Answer 3

You are not correctly applying distributive law, indeed for a simpler case we have

$$(a+a)(b+b)=a(b+b)+a(b+b)=ab+ab+ab+ab=4ab$$

from here you can derive correctly the result.

As an alternative, as you noticed, in a more direct way

$$(a+a+a)(b+b+b)=(3a)(3b)=3\cdot3\cdot ab=9ab$$

Answer 4

A bit of Geometry.

Draw a rectangle with base $3a$ and height $3b.$

Total area= $(3a)(3b).$

Let the total area be compososed of elementary rectangle units of width $a$ and of height $b$, i.e. of area $ab$ each.

How many do you count?

Answer 5

Try this for $a = b = 1$. Then $(3a) \cdot (3b) = 9 = 9(ab)$, but $3(ab) = 3$.