Bubble is a comic character. In Romanian, his name also has a trivial connotation.
Everyone knows that multiplication is distributive over addition...
$$a\cdot (b+c)=a\cdot b+a\cdot c \tag{$\cdot$ / +}$$
Here we imply a certain "priority" of multiplication over addition, so as not to write excessively pedantically $a\cdot (b+c)=(a\cdot b)+(a\cdot c)$.
What if addition were distributive to multiplication ?
A very narrow-minded math teacher would compare a student who thinks like this to Bubble. This would be written
$$a+b\cdot c=(a+b)\cdot (a+c) \tag{+/$\cdot$}$$
or, excessively pedantically, like this $a+(b\cdot c)=(a+b)\cdot (a+c).$
Let's not forget that it is possible for two operations to be mutually distributive of each other e.g. in the case of Boolean algebras.
I thought about a similar situation when I was developing initial assessment tests. I gave a problem like this to 8th graders :
Let $a,\;b,\;c$ be real numbers that satisfy
$a+b+c=1 \tag{1}$
Prove that $a+b\cdot c=(a+b)(a+c) \tag{2}$
There are many ways to demonstrate this, from a complete lack of reaction to the statement, to the most complicated calculations. Which? would be the shortest, and therefore most elegant, demonstration?? I would say that this would be :
$a+b\cdot c=a\cdot 1+bc \underset{(1)}{=}a(a+b+c)+bc=a^2+ab+ac+bc=a(a+b)+c(a+b)=(a+b)(a+c)$
The problem in the image is related to the same context.
Solution CiP
$1-ab-bc-ca\overset{(1)}{=}(a+b+c)^2-ab-bc-ca=a^2+b^2+c^2=ab+bc+ca=$
$=\frac{a^2+2ab+b^2}{2}+\frac{b^2+2bc+c^2}{2}+\frac{c^2+2ca+a^2}{2}=$
$=\underline{\frac{(a+b)^2}{2}+\frac{(b+c)^2}{2}+\frac{(c+a)^2}{2}}=$
$\overset{a+b=1-c}{\underset{and \;the\; others}{=}}\frac{(1-c)^2}{2}+\frac{(1-a)^2}{2}+\frac{(1-b)^2}{2}=\underline{\left (\frac{1-c}{\sqrt{2}}\right )^2+\left ( \frac{1-a}{\sqrt{2}}\right )^2+\left( \frac{1-b}{\sqrt{2}}\right )^2}$
$\blacksquare$
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