Math Problem Statement

8e f⁶ - 125g³

Solution

The expression you've provided is 8ef6125g38e f^6 - 125g^3. This is a difference of two terms, and we can factor it using the difference of cubes formula. The difference of cubes formula is:

a3b3=(ab)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2)

To factor 8ef6125g38e f^6 - 125g^3, let's first recognize the cube forms:

(2f2)3(5g)3(2f^2)^3 - (5g)^3

Now, apply the difference of cubes formula:

8ef6125g3=(2f25g)((2f2)2+(2f2)(5g)+(5g)2)8e f^6 - 125g^3 = \left( 2f^2 - 5g \right)\left( \left(2f^2\right)^2 + \left(2f^2\right)(5g) + (5g)^2 \right)

Simplifying the expression inside the second parentheses:

=(2f25g)(4f4+10f2g+25g2)= (2f^2 - 5g) \left( 4f^4 + 10f^2g + 25g^2 \right)

Thus, the factored form is:

(2f25g)(4f4+10f2g+25g2)\boxed{(2f^2 - 5g)(4f^4 + 10f^2g + 25g^2)}

Would you like more details on this factorization process, or do you have any questions?


Here are 5 related questions:

  1. How do you factor the sum of cubes?
  2. What are some examples of polynomials that can be factored using the difference of squares?
  3. Can all polynomials be factored using standard methods?
  4. What is the relationship between factoring and solving polynomial equations?
  5. How can factoring be used to simplify complex algebraic expressions?

Tip: When factoring polynomials, it's helpful to recognize patterns such as difference of squares or cubes, which can greatly simplify the process.

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Math Problem Analysis

Mathematical Concepts

Algebra
Polynomials
Factoring
Difference of Cubes

Formulas

Difference of cubes formula

Theorems

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Suitable Grade Level

Grades 10-12