To show how this answer comes about, let us examine what would normally happen if we tried to expand the parentheses. 1225 = 5^2 \cdot 7^2$, therefore the sum of factors is $ (1+5+25)(1+7+49) = 1767$. We note that as and can be any two numbers, this is a formula that applies to any expression that is a difference of two cubes. Let us demonstrate how this formula can be used in the following example. A simple algorithm that is described to find the sum of the factors is using prime factorization.
Using the fact that and, we can simplify this to get. Specifically, we have the following definition. In other words, is there a formula that allows us to factor? Just as for previous formulas, the middle terms end up canceling out each other, leading to an expression with just two terms. Sometimes, it may be necessary to identify common factors in an expression so that the result becomes the sum or difference of two cubes. If we do this, then both sides of the equation will be the same. Now, we have a product of the difference of two cubes and the sum of two cubes. As we can see, this formula works because even though two binomial expressions normally multiply together to make four terms, the and terms in the middle end up canceling out. 94% of StudySmarter users get better up for free. Let us see an example of how the difference of two cubes can be factored using the above identity.
This can be quite useful in problems that might have a sum of powers expression as well as an application of the binomial theorem. We also note that is in its most simplified form (i. e., it cannot be factored further). Since we have been given the value of, the left-hand side of this equation is now purely in terms of expressions we know the value of. Example 1: Finding an Unknown by Factoring the Difference of Two Cubes. Let us investigate what a factoring of might look like. This allows us to use the formula for factoring the difference of cubes. This result is incredibly useful since it gives us an easy way to factor certain types of cubic equations that would otherwise be tricky to factor. This question can be solved in two ways. In order for this expression to be equal to, the terms in the middle must cancel out. We might wonder whether a similar kind of technique exists for cubic expressions. These terms have been factored in a way that demonstrates that choosing leads to both terms being equal to zero. Are you scared of trigonometry? Therefore, we can rewrite as follows: Let us summarize the key points we have learned in this explainer.
Use the sum product pattern. Example 5: Evaluating an Expression Given the Sum of Two Cubes. Therefore, factors for. Edit: Sorry it works for $2450$. By identifying common factors in cubic expressions, we can in some cases reduce them to sums or differences of cubes. Try to write each of the terms in the binomial as a cube of an expression. An amazing thing happens when and differ by, say,. Thus, the full factoring is. This factoring of the difference of two squares can be verified by expanding the parentheses on the right-hand side of the equation. The difference of two cubes can be written as.
Thus, we can apply the following sum and difference formulas: Thus, we let and and we obtain the full factoring of the expression: For our final example, we will consider how the formula for the sum of cubes can be used to solve an algebraic problem. Substituting and into the above formula, this gives us. We can combine the formula for the sum or difference of cubes with that for the difference of squares to simplify higher-order expressions. Gauthmath helper for Chrome. It can be factored as follows: We can additionally verify this result in the same way that we did for the difference of two squares. This is because each of and is a product of a perfect cube number (i. e., and) and a cubed variable ( and). Check the full answer on App Gauthmath. Still have questions? Suppose, for instance, we took in the formula for the factoring of the difference of two cubes. Note that although it may not be apparent at first, the given equation is a sum of two cubes. Point your camera at the QR code to download Gauthmath.
We can find the factors as follows. Now, we recall that the sum of cubes can be written as. Specifically, the expression can be written as a difference of two squares as follows: Note that it is also possible to write this as the difference of cubes, but the resulting expression is more difficult to simplify. Example 4: Factoring a Difference of Squares That Results in a Product of a Sum and Difference of Cubes. We might guess that one of the factors is, since it is also a factor of. Before attempting to fully factor the given expression, let us note that there is a common factor of 2 between the terms. The sum or difference of two cubes can be factored into a product of a binomial times a trinomial. Given that, find an expression for. To see this, let us look at the term. Let us continue our investigation of expressions that are not evidently the sum or difference of cubes by considering a polynomial expression with sixth-order terms and seeing how we can combine different formulas to get the solution. If we also know that then: Sum of Cubes.
Unlimited access to all gallery answers. Since the given equation is, we can see that if we take and, it is of the desired form. Use the factorization of difference of cubes to rewrite. In the following exercises, factor. However, it is possible to express this factor in terms of the expressions we have been given. Example 2: Factor out the GCF from the two terms. Good Question ( 182). Please check if it's working for $2450$.
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