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Remedial Algebra I25 chapters | 248 lessons | 1 flashcard set

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Lesson Transcript

Instructor:
*Yuanxin (Amy) Yang Alcocer*

Amy has a master's degree in secondary education and has taught math at a public charter high school.

Watch this video lesson so you can identify perfectly cubed equations. Learn about the formula that helps you factor them and then to solve them. See how easy it is to use these formulas.

A **perfectly cubed equation** is an equation where you have a value cubed minus or plus another valued cubed. In algebra terms, we say that a perfectly cubed equation will be of the form *a*^3 - *b*^3 = 0 or *a*^3 + *b*^3 = 0. An example of a perfectly cubed equation is *x*^3 - *y*^3 = 0.

8*x*^3 + 64 = 0 is another example. Do you see how this last equation is a cubic equation? This one is a bit tricky, but if we rewrite it in terms of cubes, you will see how this is also a perfectly cubed equation. We can rewrite the 8 as 2^3, and we can rewrite the 64 as 4^3 to get 2^3**x*^3 + 4^3 = 0. Because the 2 and the *x* are both cubed, we can combine them together to get (2*x*)^3 + 4^3 = 0. Now do you see how we have one value cubed plus another value cubed?

Another way you can think of cubic equations is to just think of two cubes and what it takes to find the volumes of them. Remember that a cube's volume is calculated by multiplying its length, width and height. Since they are all the same, you just need to multiply one side together three times. You end up cubing one of the sides, *s*^3. So, a perfectly cubed equation will have the volume of one cube minus or plus the volume of another cube. You will have an exponent of 3 on both the first term and the second term once you have rewritten your equation so that it includes the exponents.

Why do you need to know how to solve these types of equations? Well, you will see these types of equations on your tests and in physics problems. You will know how to solve them easily with the formulas you will learn in this video lesson. So, let's keep going.

We have one formula when we have a minus in between the cubes. We call the minus form the **difference of cubes**. The formula for the minus form is this: *a*^3 - *b*^3 = (*a* - *b*)(*a*^2 + *ab* + *b*^2).

The plus form is called the **sum of cubes**, and the formula for this form is this: *a*^3 + *b*^3 = (*a* + *b*)(*a*^2 - *ab* + *b*^2).

Looking at these two forms, you can see that they are practically the same except for the minus and plus signs. Try to connect these signs to the signs you see in your equation. The first set of parentheses is our two values with the same sign as our original equation. If our original equation has a plus, then this will be a plus. If it is a minus, then this will be a minus, too.

The next set of parentheses starts out with our first value squared. Then it will be the opposite sign of what is in our original equation. So, if our original is a plus, then this will be a minus. If it is a minus, then this will be a plus. Next comes our two values multiplied by each other. Then we have a plus for our second value squared. The last term is always added.

Now let's see these two formulas in action.

Say we want to solve *x*^3 - *y*^3 = 0, which is a difference of cubes. We first realize that this is a difference of cubes because of the minus, so we will use the difference of cubes formula, which is *a*^3 - *b*^3 = (*a* - *b*)(*a*^2 + *ab* + *b*^2). Next we need to figure out which is *a* and which is *b*.

For our first term, we see an *x* being cubed, so that tells me that our *a* is *x*. Our second term is a *y* cubed, so our *b* is *y*. Now I can simply plug these values into our formula and solve. I am plugging in *x* for *a* and *y* for *b*. Let's see what we get.

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

Now we go ahead and solve our first set of parentheses and second set of parentheses to find our answers. We set both equal to 0 to find our answers.

*x* - *y* = 0 and *x*^2 + *xy* + *y*^2 = 0

We are solving for *x*. So, our first set of parentheses gives us *x* = *y* when we move the *y* over by adding it to both sides. To solve our second set of parentheses, we need to use what we know about solving quadratic equations. This particular equation requires the use of the quadratic formula to help us. We remember that the quadratic formula is *x* = (-*b* +/- sqrt (*b*^2 - 4*ac*))/2*a*.

Comparing this to our equation, I see that my *a* is 1, my *b* is *y* and my *c* is *y*^2. I go ahead and plug these values into my quadratic formula and this is what I get:

*x* = (-*y* +/- sqrt (*y*^2 - 4 * 1 * *y*^2))/2 * 1

Evaluating this equation, I get this:

*x* = (-*y* +/- sqrt (*y*^2 - 4*y*^2))/2, which becomes *x* = (-*y* +/- sqrt (-3*y*^2))/2

Hmm. I think I've reached the end. I see that inside my square root I will end up with a negative number. So, that tells me that there are no real solutions here. So, my only answer is *x* = *y*.

Now what about the sum of cubes? Let's solve the equation 8*x*^3 + 64 = 0 to see what happens. I use the formula for the sum of cubes, which is *a*^3 + *b*^3 = (*a* + *b*)(*a*^2 - *ab* + *b*^2). I can't use this formula right away because I need to rewrite my equation so that I can see which values are being cubed. I analyze my numbers and I see that my 8 can be rewritten as 2^3 and my 64 can be rewritten as 4^3.

My 2^3 and my *x*^3 can be combined to (2*x*)^3, so I can rewrite my equation to become (2*x*)^3 + 4^3 = 0. Now I can see clearly which values are being cubed. My *a* is 2*x* and my *b* is 4. Now I can plug these into my formula. Doing so, I get this:

(2*x*)^3 + 4^3 = (2*x* + 4)((2*x*)^2 - 2*x**4 + 4^2)

This evaluates to (2x + 4)(4x^2 - 8x + 16).

I do the same as I did with the difference of cubes to continue solving. I set both sets of parentheses equal to 0.

*2x + 4 = 0* and *4x^2 - 8x + 16 = 0*

Solving the first by subtracting 4 from both sides and then dividing by 2, I get x = -2.

To solve the second, I need to use what I know about solving quadratics. Here again, I can use the quadratic formula. Usually with cubic equations, you will need to use the quadratic formula to solve this part. For my quadratic formula, I see that my a is 4, my b is -8, and my c is 16. Plugging these into the formula I get x = (8 +/- sqrt ((-8)^2 - 4*4*16))/2*4 which becomes x = (8 +/- sqrt (64 - 256))8. The part inside the square root becomes -192, which is a negative number. So that tells me that again I have no real solutions. So my only solution here is x = -2. And we are done!

What have we learned? We've learned that a **perfectly cubed equation** is an equation where you have a value cubed minus or plus another valued cubed. We have two different forms of perfectly cubed equations. We can have the **difference of cubes**, which is *a*^3 - *b*^3 = 0, and we can have the **sum of cubes**, which is *a*^3 + *b*^3 = 0.

To solve these types of equations, we use the formulas for each form. For the difference of cubes I use the formula *a*^3 - *b*^3 = (*a* - *b*)(*a*^2 + *ab* + *b*^2), and for the sum of cubes I use the formula *a*^3 + *b*^3 = (*a* + *b*)(*a*^2 - *ab* + *b*^2).

To continue solving, I set both sets of parentheses equal to 0 and solve for *x*. I use my algebra skills to solve the first set of parentheses. I use what I know about quadratic equations to solve the second set of parentheses. It usually requires the use of the quadratic formula. If the part inside the square root is negative, then I know that there won't be any real solutions from this second set of parentheses and my only answer is the one I got from solving the first set of parentheses.

By the end of this lesson you should be able to:

- Identify a perfectly cubed equation
- Manipulate an equation to apply the difference of cubes or sum of cubes formulas to solve

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Remedial Algebra I25 chapters | 248 lessons | 1 flashcard set

- What is a Cubic Equation? - Definition & Examples 5:23
- How to Solve Perfectly Cubed Equations 10:32
- Using the Greatest Common Factor to Solve Cubic Equations 4:01
- Grouping to Factor Cubic Equations 6:41
- Solving Cubic Equations with Integers 8:33
- Changing Radical Equations into Linear or Quadratic Equations 5:14
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