Multiplication of Powers: Solving the problem

Let's start by simplifying the second term in the complete multiplication, meaning - the fraction. We'll simplify it in two stages:

In the first stage we'll use the power law for multiplication between terms with identical bases:

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

and simplify the fraction's numerator:

$\frac{19^{35}\cdot19^{-32}}{19^3}=\frac{19^{35+(-32)}}{19^3}=\frac{19^{35-32}}{19^3}=\frac{19^3}{19^3}$

Next, we can either remember that dividing any number by itself gives 1, or use the power law for division between terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$ to get that:$\frac{19^3}{19^3}=19^{3-3}=19^0=1$

where in the last step we used the fact that raising any number to the power of 0 gives 1, meaning mathematically that:

$X^0=1$

Let's summarize this part, we got that:

$\frac{19^{35}\cdot19^{-32}}{19^3}=1$

Let's now return to the complete expression in the problem and substitute this result in place of the fraction:

$3^{-3}\cdot\frac{19^{35}\cdot19^{-32}}{19^3}=3^{-3}\cdot1=3^{-3}$

In the next stage we'll recall the power law for negative exponents:

$a^{-n}=\frac{1}{a^n}$

and apply this law to the result we got:

$3^{-3}=\frac{1}{3^3}=\frac{1}{27}$

Summarizing all the steps above, we got that:

$3^{-3}\cdot\frac{19^{35}\cdot19^{-32}}{19^3}=3^{-3}=\frac{1}{27}$

Therefore the correct answer is answer A.

We use the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$We apply it in the problem:

$Y^2+Y^6-Y^5\cdot Y=Y^2+Y^6-Y^{5+1}=Y^2+Y^6-Y^6=Y^2$When we apply the previous property to the third expression from the left in the sum, and then simplify the total expression by adding like terms.

Therefore, the correct answer is option D.

We use the formula:

$a^0=1$

$\frac{9\times3}{8^0}=\frac{9\times3}{1}=9\times3$

We know that:

$9=3^2$

Therefore, we obtain:

$3^2\times3=3^2\times3^1$

We use the formula:

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

$3^2\times3^1=3^{2+1}=3^3$

First let's note that the number 9 is a power of the number 3:

$9=3^2$

therefore we can immediately move to a unified base in the problem, in addition we'll recall the law of powers for negative exponents but in the opposite direction:

$\frac{1}{a^n} =a^{-n}$

Let's apply this to the problem:

$9^4\cdot3^{-8}\cdot\frac{1}{3}=(3^2)^4\cdot3^{-8}\cdot3^{-1}$

where in the first term of the multiplication we replaced the number 9 with a power of 3, according to the relationship mentioned earlier, and simultaneously the third term in the multiplication we expressed as a term with a negative exponent according to the aforementioned law of exponents.

Now let's recall two additional laws of exponents:

a. The law of exponents for power of a power:

$(a^m)^n=a^{m\cdot n}$

b. The law of exponents for multiplication between terms with equal bases:

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

Let's apply these two laws to the expression we got in the last stage:

$(3^2)^4\cdot3^{-8}\cdot3^{-1}=3^{2\cdot4}\cdot3^{-8}\cdot3^{-1}=3^8\cdot3^{-8}\cdot3^{-1}=3^{8+(-8)+(-1)}=3^{8-8-1}=3^{-1}$

where in the first stage we applied the law of exponents for power of a power mentioned in a', in the next stage we applied the law of exponents for multiplication of terms with identical bases mentioned in b', then we simplified the resulting expression.

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

$9^4\cdot3^{-8}\cdot\frac{1}{3}=(3^2)^4\cdot3^{-8}\cdot3^{-1} =3^{8+(-8)+(-1)}=3^{8-8-1}=3^{-1}$

Therefore the correct answer is answer b'.