Factorize 6(x^5+y^5+z^5)-5(x^2+y^2+z^2)(x^3+y^3+z^3)

  • MHB
  • Thread starter anemone
  • Start date
In summary, factorization is the process of breaking down a mathematical expression into its simplest forms or factors. It is important in mathematics because it helps us to simplify complex expressions and solve equations more easily. To factorize expressions with multiple variables, we can use techniques such as the distributive property, common factor, grouping, difference of squares, and sum or difference of cubes. The FOIL method cannot be used for factorizing expressions. Factoring differs from expanding, as it works backwards and helps us find the original expression. Factoring can also be used to solve equations by simplifying the expressions and identifying common factors.
  • #1
anemone
Gold Member
MHB
POTW Director
3,883
115
Factorize $6(x^5+y^5+z^5)-5(x^2+y^2+z^2)(x^3+y^3+z^3)$ as a product of polynomials of lower degree with integer coefficients.
 
Mathematics news on Phys.org
  • #2
anemone said:
Factorize $6(x^5+y^5+z^5)-5(x^2+y^2+z^2)(x^3+y^3+z^3)$ as a product of polynomials of lower degree with integer coefficients.
$f(x,y,z) = 6(x^5+y^5+z^5)-5(x^2+y^2+z^2)(x^3+y^3+z^3)$ is a homogenous symmetric polynomial of degree $5$. According to the theory of symmetric polynomials, it can be uniquely expressed in terms of the elementary symmetric polynomials $e_1,e_2,e_3$, where $$e_1 = x+y+z,\qquad e_2 = xy+yz+zx, \qquad e_3 = xyz.$$ To write $f(x,y,z)$ in that form, use Newton's identities, which say that if $p_k = x^k+y^k+z^k$ then $$p_1=e_1,\\ p_2 = e_1^2 - e_2,\\p_3 = e_1^3 - 3e_1e_2 + 3e_3,\\ p_k= e_1p_{k-1} - e_2p_{k-2} + e_3p_{k-3}\quad (k>3).$$ Then $$\begin{aligned}p_5& = e_1p_4 - e_2p_3 + e_3p_2\\ &= e_1(e_1p_3-e_2p_2+e_3p_1) - e_2p_3 + e_3p_2\\ &= (e_1^2-e_2)p_3 + (e_3-e_1e_2)p_2 + e_1e_3p_1\\ &=(e_1^2-e_2)(e_1^3 - 3e_1e_2 + 3e_3) + (e_3-e_1e_2)(e_1^2 - 2e_2) + e_1^2e_3\\ &= e_1^5 - 5e_1^3e_2 + 5e_1^2e_3 + 5e_1e_2^2 - 5e_2e_3.\end{aligned}$$ Therefore $$6p_5 = 6e_1^5 - 30e_1^3e_2 + 30e_1^2e_3 + 30e_1e_2^2 - 30e_2e_3.\qquad(*)$$ Next, $p_2p_3 = (e_1^2 - e_2)(e_1^3 - 3e_1e_2 + 3e_3) = e_1^5 - 5e_1^3e_2 + 3e_1^2e_3 + 6e_1e_2^2 - 6e_2e_3$, and so $$5p_2p_3 = 5e_1^5 - 25e_1^3e_2 + 15e_1^2e_3 + 30e_1e_2^2 - 30e_2e_3.\qquad(**)$$ From (*) and (**) it follows that $$f(x,y,z) = 6p_5 - 5p_2p_3 = e_1^5 - 5e_1^3e_2 + 15e_1^2e_3 = e_1^2(e_1^3 - 5e_1e_2 + 15e_3).$$ In terms of $x,y,z$, that says that $$6(x^5+y^5+z^5)-5(x^2+y^2+z^2)(x^3+y^3+z^3) = (x+y+z)^2\bigl(x^3+y^3+z^3 - 2x^2(y+z) - 2y^2(x+z) - 2z^2(x+y) + 6xyz\bigr)$$ (as you can verify by multiplying out the brackets on both sides, if you are so inclined).
 
Last edited:

1. What is factorization?

Factorization is the process of breaking down a mathematical expression into its simplest form by finding its factors. This is often done to simplify the expression and make it easier to work with.

2. How do you factorize an expression?

To factorize an expression, you can use a variety of methods such as grouping, difference of squares, or trial and error. In this case, the expression can be factored by using the difference of cubes formula.

3. What is the difference of cubes formula?

The difference of cubes formula is (a^3 - b^3) = (a - b)(a^2 + ab + b^2). It can be used to factorize expressions that have the form of x^3 - y^3.

4. Can the expression be simplified further?

Yes, the expression can be further simplified by factoring out a common term from each term in the expression. In this case, the common term is (x^2 + y^2 + z^2).

5. What are the factors of the given expression?

The factors of the given expression are (x^2 + y^2 + z^2) and (6x^3 - 5x^2 + 6y^3 - 5y^2 + 6z^3 - 5z^2).

Similar threads

Replies
1
Views
875
Replies
3
Views
837
  • General Math
Replies
11
Views
378
  • General Math
Replies
3
Views
1K
Replies
1
Views
900
Replies
9
Views
1K
  • General Math
Replies
1
Views
841
Replies
15
Views
1K
Replies
4
Views
922
  • General Math
Replies
1
Views
659
Back
Top