Simplify the Expression: y³ · y⁻⁴ · (-y)³ ÷ y⁻³

Let's start by handling the term in the multiplication that is in parentheses:

$(-y)^3$

For this, we'll recall the law of exponents for an exponent of a term in parentheses:

$(a\cdot b)^n=a^n\cdot b^n$

Accordingly, we get that:

$(-y)^3=(-1\cdot y)^3=(-1)^3\cdot y^3=-1\cdot y^3=-y^3$

We'll use this understanding in the problem and apply it to the aforementioned term:

$\frac{y^3\cdot y^{-4}\cdot(-y)^3}{y^{-3}}=\frac{y^3\cdot y^{-4}\cdot(-y^3)}{y^{-3}}=\frac{-y^3\cdot y^{-4}\cdot y^3}{y^{-3}}$

where in the first stage we used the above understanding carefully - while using parentheses, and this is in order to remember that we're dealing with multiplication (not subtraction) and then we rearranged the expression using the distributive property of multiplication while remembering that a negative coefficient means multiplying by negative one,

Next, we'll recall the law of exponents for multiplying terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

and we'll apply this law to the expression we got in the last stage:

$\frac{-y^3\cdot y^{-4}\cdot y^3}{y^{-3}}=\frac{-y^{3+(-4)+3}}{y^{-3}}=\frac{-y^{3-4+3}}{y^{-3}}=-\frac{y^2}{y^{-3}}$

where in the first stage we applied the above law of exponents to the multiplication terms (with identical bases) in the expression and in the final stage we remembered that negative one divided by negative one equals negative one.

Let's summarize the solution steps so far:

$\frac{y^3\cdot y^{-4}\cdot(-y)^3}{y^{-3}}=\frac{-y^3\cdot y^{-4}\cdot y^3}{y^{-3}} =-\frac{y^2}{y^{-3}}$

We'll continue and recall the law of exponents for dividing terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$

Let's apply this law to the expression we got in the last stage:

$-\frac{y^2}{y^{-3}}=-y^{2-(-3)}=-y^{2+3}=-y^5$

where in the first stage we applied the above law of exponents carefully, because the term in the denominator has a negative exponent and then we simplified the expression in the exponent,

Let's summarize the solution steps, we got that:

$\frac{y^3\cdot y^{-4}\cdot(-y)^3}{y^{-3}}=-\frac{y^2}{y^{-3}}=-y^5$

Therefore, the correct answer is answer A.

Note:

Let's note and emphasize that the minus sign in the final answer is not under the exponent, meaning - the exponent doesn't apply to it but only to $y$, and this is in contrast to the understanding from the beginning of the solution where the entire expression: $-y$ is under the power of 3 because it's inside parentheses that are raised to the power of 3, meaning:

$(-y)^3$.