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

Let's start by dealing with the multiplication term inside parentheses:

$(-y)^3$

For this, we must recall the rule for an exponent of a term inside parentheses:

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

This gives us:

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

We'll now apply this to the aforementioned term:

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

We rearranged the expression using the distributive property of multiplication while remembering that a negative coefficient means multiplying by negative one.

Next, we need to recall the law of exponents for multiplying terms with identical bases:

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

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

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

Iin 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 by remembering 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 that we got in the last stage:

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

In the first stage we applied the above law of exponents carefully, because the term in the denominator has a negative exponent. Then we simplified the expression in the exponent.

Let's summarize the solution steps:

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

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 $y$. This is in contrast to the beginning of the solution where, for the entire expression, $-y$ is under the power of 3 because it's inside parentheses that are raised to the power of 3. Therefore:

$(-y)^3$.