Why Study Algebra? – Special Products – The Sum and Difference of Two Cubes

There’s two special products in algebra that ought to have mentioning: the sum and difference of two cubes. Even though the quadratics are usually prevalent, the cubes in addition to greater order polynomials find their devote many interesting applications. For this reason, understanding how to factor x^3   y^3 and x^3 – y^3 deserve some attention. Let’s explore them here.

For the sum situation, that’s x^3   y^3, we factor this as (x   y)(x^2 – xy   y^2). Put the twelve signs correspond within the first factor there’s however a poor term, namely -xy within the second. This can be really the reply to remembering the factorization. Keep in mind that for the sum situation, the first factor is (x   y) the second factor should have a poor. As is available x^3 and y^3, the final terms within the second factor needs to be positive. Because we wish this mix terms to cancel, we have to possess a negative for the other term.

For the difference situation, that’s x^3 – y^3, we factor this as (x – y)(x^2   xy   y^2). Put the twelve signs correspond within the first factor but all signs within the second are positive. That many, is essential to remembering the factorization. Keep in mind that for the difference situation, the first factor is (x – y) the second factor has all positives. This insures this mix terms cancel and we’re playing just x^3 – y^3.

To know the exposition above, let’s really multiply the sum situation out (the main difference scenario is entirely similar). We’ve

(x   y)(x^2 – xy   y^2) = x(x^2 -xy   y^2)   y(x^2 -xy   y^2). Notice generate income have attempted personally the distributive property to part ways this multiplication. Inside the finish, that maybe what this property does. The initial yields x^3 – x^2y   xy^2 along with the second yields, x^2y – xy^2   y^3. (Observe while using the commutative property of multiplication, that’s yx^2 = x^2y). Adding the 2 pieces together, we’ve -x^2y   xy^2 cancel with x^2y – xy^2. Thus all we’re playing is x^3   y^3.

What becomes a little more challenging may be the factoring in the perfect cube along with a number, is another perfect cube. Thus x^3   8. Once we write this as x^3   2^3, we percieve that folks can factor this into (x   2)(x^2 – 2x   4). Once we consider 2 as y, you need to see this exactly matches everything you have recently done. To make certain this is very apparent, consider x^3 – 27. Since 27 is equivalent to 3^3, that is therefore an ideal cube, we’re able to apply everything you just learned and write x^3 – 27 = x^3 – 3^3 = (x – 3)(x^2   3x   9).