I think the operating rule is how to deal with a power raised to another power, instead of the multiplication of the same base raised to two different powers. That says

(x^y)^z = x^(y*z)

...or we multiply the two exponents to get the final simplified power of x (because there's z copies of x^y getting multiplied together). Using numbers, (x^3)^4 would simplify to x^(3*4) = x^12.

With understood multiplication, we may want to look at it more algebraically (abc)^2 = abcabc. We can then use the commutative property of real numbers with multiplication to rearrange the factors into aabbcc, which gives us Mona's answer immediately. I think this intermediate step can be important to understand, even if it's technical. I think if we follow Mona's reasoning backward, my explanation is equivalent (minus the very good note concerning exponents and multiplication, abc = (abc)^1).

abc is being used as a factor twice therefore according to the laws of exponents when we are multiplying exponents we add them. In this case abc times abc would be a^2b^2c^2. abcxabc each variable is raised to the first power so 1+1 equals 2 and therefore the exponent is equal to 2 and you do this for each variable.

## Comments

Thank you. That is an excellent way to rewrite the question. Thank you for sharing that way of presenting it with me.

With understood multiplication, we may want to look at it more algebraically (abc)^2 = abcabc. We can then use the commutative property of real numbers with multiplication to rearrange the factors into aabbcc, which gives us Mona's answer immediately. I think this intermediate step can be important to understand, even if it's technical. I think if we follow Mona's reasoning backward, my explanation is equivalent (minus the very good note concerning exponents and multiplication, abc = (abc)^1).