Factoring $x^6-64 Y^3$: A Step-by-Step Guide

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Hey guys! Today, we're diving into a fun algebra problem: factoring the expression $x^6 - 64y^3$. This might look intimidating at first, but don't worry, we'll break it down step by step so it's super easy to understand. Factoring expressions like this is a fundamental skill in algebra, and mastering it will help you tackle more complex problems later on. So, let's get started and make sure you've got this down pat!

Understanding the Problem: The Expression $x^6 - 64y^3$

So, what exactly are we dealing with here? We've got the expression $x^6 - 64y^3$. This looks like a difference of cubes, but with a little twist. The key is to recognize that both $x^6$ and $64y^3$ can be expressed as perfect cubes. Let's break it down:

  • x^6$ can be seen as $(x^2)^3$ because when you raise a power to another power, you multiply the exponents (2 * 3 = 6).

  • 64y^3$ can be seen as $(4y)^3$ because 64 is 4 cubed (4 * 4 * 4 = 64).

Now, our expression looks like $(x2)3 - (4y)^3$. See? It's a difference of cubes! Understanding this is the first crucial step because it allows us to use a specific factoring formula.

Why is recognizing this form so important? Well, the difference of cubes has a neat little formula that makes factoring much easier. If we didn't recognize it, we might be tempted to try other, more complicated methods, which could lead us down the wrong path. By identifying the structure of the expression, we can choose the most efficient and correct way to factor it. This is a common theme in algebra – spotting patterns and structures simplifies problem-solving. It's like having a secret code that unlocks the solution!

The Difference of Cubes Formula: Our Secret Weapon

Alright, now that we've identified our expression as a difference of cubes, let's bring out our secret weapon: the difference of cubes formula. This formula is essential for factoring expressions in the form of $a^3 - b^3$, and it looks like this:

a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2)

This formula might seem a bit intimidating at first glance, but it's actually quite straightforward once you understand what each part represents. The left side of the equation, $a^3 - b^3$, is the difference of two cubes. The right side is the factored form, which consists of two factors:

  1. (a - b)$: This is the difference of the cube roots of the original terms.

  2. (a^2 + ab + b^2)$: This is a quadratic expression that's a bit trickier to remember, but it's **crucial** for completing the factorization. It consists of the square of the first term (a^2), the product of the two terms (ab), and the square of the second term (b^2).

Why is this formula so powerful? It allows us to take a seemingly complex expression and break it down into simpler factors. This is incredibly useful in algebra because factored expressions are easier to work with when solving equations, simplifying fractions, or analyzing functions. Think of it like dismantling a complicated machine into its individual parts – once you have the parts, you can understand how the whole thing works.

Applying the Formula to $x^6 - 64y^3$

Okay, let's put our secret weapon to work! We know our expression is $(x2)3 - (4y)^3$, and we know the difference of cubes formula is $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. Now, we just need to figure out what 'a' and 'b' are in our case. Comparing our expression to the formula, we can see:

  • a=x2a = x^2

  • b=4yb = 4y

Now that we've identified 'a' and 'b', we can plug them into the formula. This is like following a recipe – we have the ingredients (a and b) and the instructions (the formula), so we just need to put them together correctly.

So, let's substitute: $

(x2)3−(4y)3=(x2−4y)((x2)2+(x2)(4y)+(4y)2)(x^2)^3 - (4y)^3 = (x^2 - 4y)((x^2)^2 + (x^2)(4y) + (4y)^2)

See? We've just replaced 'a' and 'b' with their values. But we're not done yet! We need to simplify the expression on the right side. This is where our algebraic skills come into play. We need to remember our order of operations (PEMDAS/BODMAS) and carefully simplify each term.

Simplifying the Factored Expression

Now that we've substituted 'a' and 'b' into the formula, we have:

(x2−4y)((x2)2+(x2)(4y)+(4y)2)(x^2 - 4y)((x^2)^2 + (x^2)(4y) + (4y)^2)

Let's simplify the second factor step by step:

  1. (x^2)^2 = x^4$ (Remember, when you raise a power to another power, you multiply the exponents).

  2. (x2)(4y)=4x2y(x^2)(4y) = 4x^2y

  3. (4y)^2 = 16y^2$ (We need to square both the 4 and the y).

Now, let's put it all together. Our simplified expression looks like this:

(x2−4y)(x4+4x2y+16y2)(x^2 - 4y)(x^4 + 4x^2y + 16y^2)

This is the completely factored form of our original expression! We've taken a complex expression and broken it down into two simpler factors. This is a fantastic achievement, guys, because it shows we've mastered the difference of cubes formula and know how to apply it effectively.

Checking Our Work: A Crucial Step

Before we celebrate too much, there's one crucial step we need to take: checking our work. In math, it's always a good idea to verify your answer to make sure you haven't made any mistakes. How can we check our factored expression? The easiest way is to multiply the factors back together and see if we get our original expression.

So, let's multiply $(x^2 - 4y)(x^4 + 4x^2y + 16y^2)$. This will involve using the distributive property (or the FOIL method) to multiply each term in the first factor by each term in the second factor. It might seem a bit tedious, but it's worth it for the peace of mind.

When we multiply it out, we get:

x2(x4+4x2y+16y2)−4y(x4+4x2y+16y2)x^2(x^4 + 4x^2y + 16y^2) - 4y(x^4 + 4x^2y + 16y^2)

=x6+4x4y+16x2y2−4x4y−16x2y2−64y3= x^6 + 4x^4y + 16x^2y^2 - 4x^4y - 16x^2y^2 - 64y^3

Notice that the $4x^4y$ and $ - 4x^4y$ terms cancel out, and the $16x2y2$ and $ - 16x2y2$ terms also cancel out. This leaves us with:

x6−64y3x^6 - 64y^3

This is exactly our original expression! So, we can confidently say that our factored form is correct. This check step is a fantastic way to build confidence in your answer and catch any potential errors.

Conclusion: Mastering Factoring

Awesome job, guys! We've successfully factored the expression $x^6 - 64y^3$ using the difference of cubes formula. We started by recognizing the expression as a difference of cubes, then we applied the formula, simplified the result, and even checked our work to make sure we were correct. This is a fantastic example of how understanding algebraic formulas and techniques can help us solve complex problems.

Factoring is a fundamental skill in algebra, and mastering it will open doors to more advanced topics. By practicing these techniques and understanding the underlying concepts, you'll become a more confident and skilled mathematician. Keep up the great work, and remember to always check your answers!