How to Solve Perfectly Cubed Equations

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  • 0:02 Perfectly Cubed Equation
  • 2:10 The Formulas
  • 3:37 The Difference of Cubes
  • 6:25 The Sum of Cubes
  • 9:04 Lesson Summary
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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.

Perfectly Cubed Equation

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.

8x^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 (2x)^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.

The Formulas

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.

The Difference of Cubes

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 - 4ac))/2a.

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

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