Which method is easier and why?

In summary, the problem is to factor x^6 - y^6 as a difference of squares and then as a difference of cubes. The conversation discusses the two methods and their ease of use, with the conclusion that the difference of squares method may be simpler in subsequent steps. Two possible factorizations are also provided, one using the difference of squares method and the other using the difference of cubes method.
  • #1
mathdad
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1
I found the following problem in Section 1.3 of my Precalculus textbook by David Cohen.

Factor x^6 - y^6 as a difference of squares and then as a difference of cubes.

1. Which method is easier and why?

2. Can someone get me started?
 
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  • #3
Excellent!
 
  • #4
RTCNTC said:
I found the following problem in Section 1.3 of my Precalculus textbook by David Cohen.

Factor x^6 - y^6 as a difference of squares and then as a difference of cubes.

1. Which method is easier and why?

2. Can someone get me started?

you can apply either way. but next steps become simpler if you apply difference of square

$x^6-y^6= (x^3+y^3)(x^3-y^3)$ 1st term is sum of cubes and second is difference of cubes and you get

$x^6-y^6= (x^3+y^3)(x^3-y^3)= (x+y)(x^2-xy+y^2)(x-y)(x^3+xy+y^2)$

had you chosen difference of cubes you would have got

$x^6-y^6 = (x^2-y^2)(x^4+x^2y^2+y^4)$

factoring $x^4+x^2y^2+y^4 = (x^4+2x^2y^2+y^4)- x^2y^2 = (x^2+y^2)^2 - (xy)^2 = (x^2+y^2+xy)(x^2+y^2-xy)$

above may not be straight forward but doable
 
  • #5
kaliprasad said:
you can apply either way. but next steps become simpler if you apply difference of square

$x^6-y^6= (x^3+y^3)(x^3-y^3)$ 1st term is sum of cubes and second is difference of cubes and you get

$x^6-y^6= (x^3+y^3)(x^3-y^3)= (x+y)(x^2-xy+y^2)(x-y)(x^3+xy+y^2)$

had you chosen difference of cubes you would have got

$x^6-y^6 = (x^2-y^2)(x^4+x^2y^2+y^4)$

factoring $x^4+x^2y^2+y^4 = (x^4+2x^2y^2+y^4)- x^2y^2 = (x^2+y^2)^2 - (xy)^2 = (x^2+y^2+xy)(x^2+y^2-xy)$

above may not be straight forward but doable

Nice work in every sense.
 

1. What is the difference between cubes and squares?

The main difference between cubes and squares is their dimensionality. Squares are two-dimensional shapes with four equal sides and four right angles, while cubes are three-dimensional shapes with six equal square faces. Another difference is that squares have a length and width, while cubes have a length, width, and height.

2. How do you calculate the difference of cubes?

The difference of cubes refers to the algebraic expression (a^3 - b^3). To calculate this difference, you can use the formula (a - b)(a^2 + ab + b^2). Alternatively, you can also expand the expression (a^3 - b^3) using the distributive property, which results in a^3 - b^3 = a^3 - ab^2 - a^2b + b^3.

3. What is the difference of squares formula?

The difference of squares formula is (a^2 - b^2) = (a - b)(a + b). This formula is used to factorize the difference of two perfect squares, resulting in a product of two binomials.

4. What are the applications of the difference of cubes and squares?

The difference of cubes and squares has various applications in mathematics and other fields. In algebra, it is used to factorize polynomial expressions, simplify equations, and solve equations. In geometry, it is used to calculate the volume of a cube or the area of a square. In statistics, it is used to calculate the difference between two data points. In physics, it is used to calculate the difference in energy between two states.

5. Can you give an example of the difference of cubes and squares?

Sure, an example of the difference of cubes is (5^3 - 2^3) = (5 - 2)(5^2 + 5*2 + 2^2) = (3)(25 + 10 + 4) = 117. An example of the difference of squares is (9^2 - 6^2) = (9 - 6)(9 + 6) = (3)(15) = 45.

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