Unlock The Factored Form: 125a^6 - 64 Explained

Unlock the Factored Form: $125a^6 - 64$ Explained

Hey math enthusiasts! Ever stared at an expression like $125a^6 - 64$ and wondered how on earth you're supposed to break it down? Don't worry, you're not alone. Factoring can sometimes feel like solving a mystery, but with the right tools and a little practice, you'll be decoding these algebraic puzzles in no time. Today, we're going to dive deep into finding the factored form of $125a^6 - 64$. This isn't just about getting the right answer; it's about understanding the why behind the steps. We'll explore the algebraic identities that make this process a breeze and walk through the solution step-by-step, so you can tackle similar problems with confidence. Get ready to demystify the world of factoring and boost your algebraic skills!

Understanding the Structure: Difference of Cubes!

When we look at the expression $125a^6 - 64$, the first thing a seasoned mathematician might notice is its structure. It strongly resembles the pattern of a difference of cubes. Remember that handy algebraic identity: $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$? This identity is your best friend when dealing with expressions that can be broken down into the difference of two perfect cubes. To use this, we need to identify what our 'x' and 'y' are.

Let's break down $125a^6$: We know that $125$ is $5^3$. And for $a^6$, we can think of it as $(a^2)^3$. Why $(a^2)^3$? Because when you raise a power to another power, you multiply the exponents, so $(a^2)^3 = a^{2 imes 3} = a^6$. Therefore, $125a^6$ is the perfect cube of $5a^2$. So, our 'x' in the difference of cubes formula is $5a^2$.

Now, let's look at $64$. We know that $64$ is $4^3$. So, our 'y' in the difference of cubes formula is $4$.

With $x = 5a^2$ and $y = 4$, we can now plug these into the difference of cubes formula:

$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$

Substituting our values:

$(5a^2)^3 - (4)^3 = (5a^2 - 4)((5a^2)^2 + (5a^2)(4) + (4)^2)$

This gives us the first part of our factored form: $(5a^2 - 4)$. Now, we need to expand the second part of the formula.

Expanding the Second Factor: The Key to the Complete Solution

We've successfully identified the first factor as $(5a^2 - 4)$ using the difference of cubes formula. Now, the crucial step is to correctly expand and simplify the second part of the formula: $(x^2 + xy + y^2)$.

Let's substitute our values for x and y again: $x = 5a^2$ and $y = 4$.

We need to calculate:

• $x^2$: This is $(5a^2)^2$. Remember that when you square a term with a coefficient and a variable raised to a power, you square both the coefficient and the variable part. So, $(5a^2)^2 = 5^2 imes (a^2)^2 = 25 imes a^{2 imes 2} = 25a^4$.
• $xy$: This is $(5a^2)(4)$. Simply multiply the numbers together: $5 imes 4 imes a^2 = 20a^2$.
• $y^2$: This is $(4)^2$. $4 imes 4 = 16$.

Now, we combine these parts according to the formula $(x^2 + xy + y^2)$:

$25a^4 + 20a^2 + 16$

So, the complete factored form of $125a^6 - 64$ is the product of our first factor and this expanded second factor:

$(5a^2 - 4)(25a^4 + 20a^2 + 16)$

This is the final answer. It's important to ensure that the second factor cannot be factored further using simple methods. In this case, $25a^4 + 20a^2 + 16$ doesn't readily break down into simpler binomials with real coefficients.

Analyzing the Options: Spotting the Correct Answer

Now that we've worked through the solution, let's look at the given options to see which one matches our result. This is a critical step in multiple-choice questions to confirm your understanding and catch any potential errors.

• A. $(25 a^2+16) ed{(25 a^4+20 a^2-4)}$: This option has the correct second part but an incorrect first part. It also incorrectly signs the last term in the second factor.
• B. $(5 a^2-4) ed{(25 a^4+20 a^2+16)}$: This option perfectly matches our derived factored form. The first factor $(5a^2 - 4)$ is correct, and the second factor $(25a^4 + 20a^2 + 16)$ is also correct, reflecting the expansion of $x^2 + xy + y^2$ from our difference of cubes formula.
• C. $(25 a^2-16) ed{(25 a^4-20 a^2+16)}$: This option starts with a difference of squares on the first term and has incorrect signs and coefficients in the second factor.
• D. $(5 a^2-4) ed{(25 a^4-20 a^2)}$: While the first factor is correct, the second factor is incomplete and has an incorrect sign. It seems to have missed the '+16' term and also incorrectly signed the middle term.

By comparing our calculated factored form with the given choices, it's clear that option B is the correct answer. This reinforces the steps we took using the difference of cubes identity.

The Power of Algebraic Identities: Beyond Just Memorization

Understanding and applying algebraic identities is a cornerstone of mastering algebra. The difference of cubes formula, $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$, is just one of many powerful tools in your mathematical arsenal. Others include the difference of squares ($a^2 - b^2 = (a-b)(a+b)$) and the sum of cubes ($x^3 + y^3 = (x + y)(x^2 - xy + y^2)$). Recognizing these patterns in expressions like $125a^6 - 64$ can transform a seemingly daunting problem into a straightforward application of a known rule.

It's not enough to simply memorize these formulas; the real skill lies in identifying when and how to apply them. This involves looking at the structure of the given expression. Are there two terms? Are they being subtracted (or added for sum of cubes)? Can each term be expressed as a cube (or square) of another expression? In our case, $125a^6$ is $(5a^2)^3$ and $64$ is $(4)^3$. This recognition is the key that unlocks the problem. Once identified, you meticulously substitute the base terms into the formula.

Pay close attention to the signs and coefficients within the formula. For the difference of cubes, the second factor is always a trinomial ($x^2 + xy + y^2$) with all positive terms. The sum of cubes formula has a similar structure but with a negative middle term in the trinomial: $x^2 - xy + y^2$. Mistakes often creep in during the expansion and simplification of this second factor, so double-checking $(5a^2)^2$, $(5a^2)(4)$, and $(4)^2$ is crucial. We found these to be $25a^4$, $20a^2$, and $16$, respectively, leading to the correct trinomial $25a^4 + 20a^2 + 16$.

Practice is your best ally here. The more expressions you factor, the quicker you'll become at spotting these patterns and applying the identities correctly. Try factoring different forms of polynomial expressions, and don't shy away from checking your work by multiplying the factors back together to see if you arrive at the original expression. This verification step is invaluable for solidifying your understanding and ensuring accuracy.

Conclusion: Mastering Algebraic Factoring

Finding the factored form of $125a^6 - 64$ is a fantastic exercise in recognizing and applying the difference of cubes identity. By carefully identifying $125a^6$ as $(5a^2)^3$ and $64$ as $(4)^3$, we were able to use the formula $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$ to arrive at the correct factored form: $(5a^2 - 4)(25a^4 + 20a^2 + 16)$.

This process highlights the elegance and efficiency of algebraic identities in simplifying complex expressions. Remember, the key is to look for patterns – specifically, the difference of two perfect cubes in this instance. Always double-check your substitutions and expansions, especially when dealing with exponents and coefficients. This meticulous approach will help you avoid common errors and build confidence in your algebraic abilities.

For further exploration into the fascinating world of algebra and factoring, you can always consult reliable resources. A great place to start is Khan Academy, which offers comprehensive lessons and practice exercises on a wide range of mathematical topics, including polynomial factorization. You might also find the resources at Math is Fun incredibly helpful for breaking down complex concepts in an accessible way.