Complete Factoring Of Binomials: $r^3 S^3 - T^{12}$

When we talk about factoring in mathematics, we're essentially breaking down a complex expression into simpler parts that, when multiplied together, give us the original expression. It's like taking apart a toy to see how its pieces fit together, but in reverse! The binomial we're focusing on today is $r^3 s^3 - t^{12}$. This expression, with its two terms separated by a minus sign, presents a unique opportunity for us to explore different factoring techniques. Our goal is to factor this binomial completely, meaning we want to break it down into its most basic, irreducible factors. This process often involves recognizing specific algebraic patterns, and for $r^3 s^3 - t^{12}$, we'll be looking at the difference of cubes and potentially the difference of squares if we approach it strategically.

Let's start by examining the structure of $r^3 s^3 - t^{12}$. We have two terms: $r^3 s^3$ and $t^{12}$. The first term, $r^3 s^3$, can be rewritten as $(rs)^3$. This is because when you raise a product to a power, you raise each factor to that power, and vice versa. So, $(rs)^3 = r^3 s^3$. This is a crucial observation because it immediately flags the first term as a perfect cube. Now, let's look at the second term, $t^{12}$. Can we express this as a perfect cube? Yes, we can! Remember that $(a^m)^n = a^{m imes n}$. We want to find a power $m$ such that $m imes 3 = 12$. Dividing 12 by 3, we get 4. Therefore, $t^{12} = (t^4)^3$. With these rewrites, our binomial $r^3 s^3 - t^{12}$ becomes $(rs)^3 - (t^4)^3$. This is now clearly in the form of a difference of cubes: $a^3 - b^3$, where $a = rs$ and $b = t^4$.

The general formula for factoring a difference of cubes is $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. Applying this formula to our expression, with $a = rs$ and $b = t^4$, we substitute these values into the formula. The first part of the factored form is $(a - b)$, which becomes $(rs - t^4)$. The second part is $(a^2 + ab + b^2)$. Let's break this down:

• $a^2 = (rs)^2 = r^2 s^2$
• $ab = (rs)(t^4) = rs t^4$
• $b^2 = (t^4)^2 = t^{4 imes 2} = t^8$

Putting these together, the second factor is $(r^2 s^2 + rs t^4 + t^8)$.

So, the complete factorization of $r^3 s^3 - t^{12}$ using the difference of cubes formula is $(rs - t^4)(r^2 s^2 + rs t^4 + t^8)$. At this point, we need to ask ourselves if either of these factors can be factored further. The first factor, $(rs - t^4)$, is a binomial. It's not a difference of squares (unless $r$, $s$, or $t$ are further powers), a difference of cubes, or a sum of cubes in its current form. The second factor, $(r^2 s^2 + rs t^4 + t^8)$, is a trinomial. Recognizing patterns in trinomials can be complex, but this particular trinomial doesn't immediately resemble a perfect square trinomial (which would be of the form $x^2 + 2xy + y^2$ or $x^2 - 2xy + y^2$). Given the variables and exponents, it's unlikely this trinomial can be factored further using standard algebraic techniques without introducing more complex numbers or specific contexts not provided.

An Alternative Approach: Recognizing the Difference of Squares

Interestingly, we can also factor $r^3 s^3 - t^{12}$ by first recognizing it as a difference of squares, although this requires a slightly different initial step and might lead to a different intermediate form before reaching the final factored expression. The key here is to see if both terms can be expressed as squares. We already know $r^3 s^3 = (rs)^3$. To express it as a square, we need to think about exponents. The exponent 3 isn't easily divisible by 2 to give a whole number. However, let's reconsider the original expression $r^3 s^3 - t^{12}$. We can rewrite $r^3 s^3$ as $(r^{3/2} s^{3/2})^2$, but this involves fractional exponents, which we usually try to avoid in basic factoring unless necessary. A more common approach is to recognize that if we can express both terms as squares, we can use the difference of squares formula: $a^2 - b^2 = (a - b)(a + b)$.

Let's focus on $t^{12}$. We can write $t^{12}$ as $(t^6)^2$. This is a perfect square. Now, how can we express $r^3 s^3$ as a perfect square? This is where it gets a bit tricky if we insist on only integer exponents for the base. If we're allowed fractional exponents, then $r^3 s^3 = (r^{3/2} s^{3/2})^2$. In that case, our binomial becomes $(r^{3/2} s^{3/2})^2 - (t^6)^2$. Applying the difference of squares formula, we would get $(r^{3/2} s^{3/2} - t^6)(r^{3/2} s^{3/2} + t^6)$. This form, however, is not usually considered