Simplify (a⁷/a⁵) × (a⁻²/a⁻⁴) + (a⁷)³: Complete Exponent Expression

Simplify the following expression:

$\frac{a^7}{a^5}\times\frac{a^{-2}}{a^{-4}}+(a^7)^3=$

a. Let's start by working on the first expression from the left, which is a multiplication of fractions. However, we won't perform the fraction multiplication; instead notice that in each of the fractions in the multiplication, there are terms in both the numerator and denominator with identical bases.

Therefore we'll apply the power law for division between terms with identical bases:

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

We'll apply this to our problem in the first expression from the left and apply the above power law separately to each term in the fraction multiplication:

$\frac{a^7}{a^5}\cdot\frac{a^{-2}}{a^{-4}}=a^{7-5}\cdot a^{-2-(-4)}=a^2\cdot a^{-2+4}=a^2\cdot a^2$

Note that we have an algebraic expression that is in fact the term multiplied by itself, therefore we can write the expression as a power:

$a^2\cdot a^2=(a^2)^2$

(Of course we could also apply the multiplication law between terms with identical bases to the above expression, but here we want to make use of the power of a power law, so we'll choose this way instead),

Now let's recall the power of a power law:

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

and apply it to the expression that we obtained:

$(a^2)^2=a^{2\cdot2}=a^4$

We've finished handling the first expression from the left as shown below:

$\frac{a^7}{a^5}\cdot\frac{a^{-2}}{a^{-4}}=a^4$

Remember this result and move on to the next expression.

b. For the second expression from the left, we'll once again apply the power of a power law mentioned in section a:

$(a^7)^3=a^{7\cdot3}=a^{21}$

Let's combine both solutions (a and b) into the simplified expression as follows:

$\frac{a^7}{a^5}\cdot\frac{a^{-2}}{a^{-4}}+(a^7)^3=a^4+a^{21}$

Therefore the correct answer is b.