Factoring $r^{27} - S^{30}$: A Step-by-Step Guide

Hey guys! Today, we're diving into a neat little algebra problem: factoring the expression $r^{27} - s^{30}$. This might look intimidating at first glance, but don't worry, we'll break it down into manageable steps. We'll explore how to recognize the structure of the expression, apply the difference of squares or difference of cubes formulas, and simplify to arrive at the fully factored form. So, grab your pencils, and let's get started!

Recognizing the Structure

To begin, let's recognize the structure of the expression $r^{27} - s^{30}$. Notice that both terms are perfect powers. Specifically, $r^{27}$ can be written as $(r^9)^3$ and $s^{30}$ can be written as $(s^{10})^3$. This is crucial because it allows us to use the difference of cubes formula. The difference of cubes formula states that $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.

In our case, we can consider $a = r^9$ and $b = s^{10}$. Substituting these into the difference of cubes formula, we get:

$(r^9)^3 - (s^{10})^3 = (r^9 - s^{10})((r^9)^2 + (r^9)(s^{10}) + (s^{10})^2)$

Now, let's simplify the second term:

$(r^9 - s^{10})(r^{18} + r^9s^{10} + s^{20})$

So, we have factored the expression once, but it's essential to check if we can factor it further. The term $(r^9 - s^{10})$ looks like it might be factorable if we consider other possible power representations. We can rewrite $r^9$ as $(r^3)^3$ and $s^{10}$ as $(s^{10/3})^3$, but since we want integer exponents, let's explore another route.

Alternatively, we can express the original expression $r^{27} - s^{30}$ as a difference of squares. To do this, we rewrite $r^{27}$ as $(r^{27/2})^2$ and $s^{30}$ as $(s^{15})^2$. However, having non-integer exponents isn't ideal, so let's stick with the difference of cubes approach for now.

Another approach is to think of $r^{27} - s^{30}$ as $(r^{9})^3 - (s^{10})^3$. Applying the difference of cubes factorization $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ where $a = r^9$ and $b = s^{10}$, we obtain:

$(r^9 - s^{10})((r^9)^2 + r^9s^{10} + (s^{10})^2)$ which simplifies to $(r^9 - s^{10})(r^{18} + r^9s^{10} + s^{20})$.

Now, let's consider if $r^9 - s^{10}$ can be further factored. We can write $r^9 - s^{10}$ as $(r^3)^3 - (s^{10/3})^3$, but we need integer powers. Let's consider the difference of squares: If we had something like $r^{18} - s^{20}$, we could write it as $(r^9 - s^{10})(r^9 + s^{10})$. However, we have $r^9 - s^{10}$, so let's analyze this more closely.

If we examine $r^9 - s^{10}$, we can also consider rewriting it as a difference of powers. However, without additional information, it's challenging to simplify it further using elementary factoring techniques. Therefore, we will focus on the factorization we have already achieved.

Applying the Difference of Cubes Formula

Let's revisit our application of the difference of cubes formula:

$r^{27} - s^{30} = (r^9)^3 - (s^{10})^3$

Using the formula $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ with $a = r^9$ and $b = s^{10}$, we get:

$(r^9 - s^{10})((r^9)^2 + r^9s^{10} + (s^{10})^2)$

Which simplifies to:

$(r^9 - s^{10})(r^{18} + r^9s^{10} + s^{20})$

Now, let's think about whether we can factor the term $(r^9 - s^{10})$ even further. We can express $r^9$ as $(r^3)^3$, but $s^{10}$ is not a perfect cube with an integer exponent. Similarly, we could try to express $r^9 - s^{10}$ as a difference of squares, but neither $r^9$ nor $s^{10}$ are perfect squares with integer exponents. Thus, we cannot directly apply the difference of squares or difference of cubes formulas to factor $(r^9 - s^{10})$ further using elementary methods.

So, after applying the difference of cubes formula, we have:

$r^{27} - s^{30} = (r^9 - s^{10})(r^{18} + r^9s^{10} + s^{20})$

Simplifying to the Factored Form

So far, we have $r^{27} - s^{30} = (r^9 - s^{10})(r^{18} + r^9s^{10} + s^{20})$. We need to check if we can further factor any of these terms. The term $(r^{18} + r^9s^{10} + s^{20})$ doesn't seem to fit any standard factoring patterns, like difference of squares or cubes, or perfect square trinomials. Therefore, we'll focus on $(r^9 - s^{10})$.

As we discussed before, $r^9 - s^{10}$ does not easily factor further using elementary techniques because we cannot express both terms as perfect squares or perfect cubes with integer exponents. Thus, we can conclude that the factored form we have obtained is the most simplified form achievable through basic factoring methods.

Therefore, the factored form of $r^{27} - s^{30}$ is:

$(r^9 - s^{10})(r^{18} + r^9s^{10} + s^{20})$

This is the final simplified and factored form of the given expression.

Conclusion

Alright, guys, we've successfully factored the expression $r^{27} - s^{30}$! By recognizing the structure as a difference of cubes and applying the appropriate formula, we were able to break it down into $(r^9 - s^{10})(r^{18} + r^9s^{10} + s^{20})$. Remember, the key is to look for patterns and see if you can apply known factoring formulas. Keep practicing, and you'll become a factoring pro in no time! Hope this helped, and happy factoring!