If $x=\frac{-1}{2} , y=\frac{3}{4} , z=\frac{-3}{2}$ . Find $x^3 \div y^2 z^2$ If I have Question like that :
If $ x=\frac{-1}{2} \ , \ y=\frac{3}{4} \ , \ z=\frac{-3}{2} \ $
Find a numerical Value for $$x^3 \div y^2z^2$$
First If we divide $x^3$ over $y^2$ and then multiplying the result we get $\frac{-1}{2}$
, but if I think we should first Multiply $y^2z^2$ and then we divide $x^3$ over the result , and the answer in that case is $\frac{-8}{81}$
First We notice that $$z=-2y \ \ , x=\frac{-2}{3}y$$
So $$x^3 \div y^2z^2 = (\frac{-2}{3})^3 \times y^3 \div 4\  y^4 = \frac{-8}{81} $$
For example $$6 \div 9 = \frac{2}{3}$$
And Notice if we wrote $$6 \div 9 = 6 \div 3 \times 3 = 6 \div 3^2$$
Notice $6 \div 9 = \frac{2}{3}$ but $6 \div 3 \times 3 = 6$
The write way is $$6 \div 9 =6 \div (3 \times 3) , 6 \div 3 \times 3 \neq 6 \div 9$$
And $$6\div 3^2 \neq 6 \div 3\times 3 \ \text{but} \ 6 \div 3^2 = 6 \div (3 \times 3)$$


Find a numerical Value for $$x^3 \div y^2z^2$$

I find this notation a bit ambiguous, although I would assume the intended interpretation is:
$$\frac{x^3}{y^2z^2} \tag{$*$}$$
The division operator $\div$ isn't used much in mathematics (at least not 'later on') and if you would write something like $x^3/y^2z^2$, the strict interpretation would be:
$$\frac{x^3}{y^2}z^2 \tag{$\star$}$$
because division and multiplication are at the same level and you apply them as they appear from left to right. Almost all (mathematical) software will interpret $x^3/y^2z^2$ in this way, although when written by humans, some will mean $(*)$ but this is to be avoided.
When you can use fractions, such as in $(*)$ and $(\star)$, no confusion is possible. If you're forced to write 'in line', then parentheses can be used to make sure $x^3/(y^2z^2)$ gives $(*)$, at least if that's what you mean. If you actually mean $(\star)$, then although $x^3/y^2z^2$ is technically correct, writing it as $x^3z^2/y^2$ or as $x^3z^2 \div y^2$ is a lot clearer and avoids this ambiguity and possible confusion.
