Verified Answer :

  • Identify the expression as a difference of cubes: \((3v^2)^3 - (2x^2)^3\).
  • Apply the identity \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\) to get \((3v^2 - 2x^2)(9v^4 + 6v^2x^2 + 4x^4)\).
  • Factor the expression further over the real numbers to obtain \(\boxed{(\sqrt{3}v - \sqrt{2}x)(\sqrt{3}v + \sqrt{2}x)(3v^2 - \sqrt{6}vx + 2x^2)(3v^2 + \sqrt{6}vx + 2x^2)}\).

Explanation

  1. Recognizing the Difference of Cubes

The goal is to factor the expression \(27v^6 - 8x^6\) completely. We recognize that both terms are perfect cubes: \(27v^6 = (3v^2)^3\) and \(8x^6 = (2x^2)^3\). This allows us to use the difference of cubes identity, which states that \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). In our case, letting \(a = 3v^2\) and \(b = 2x^2\), we can substitute these into the identity.

  1. Initial Factorization

Substituting \(a = 3v^2\) and \(b = 2x^2\) into the formula \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\), we get:
\[(3v^2)^3 - (2x^2)^3 = (3v^2 - 2x^2)((3v^2)^2 + (3v^2)(2x^2) + (2x^2)^2)\]
Now, we simplify the terms within the second factor:
\[(3v^2)^2 = 9v^4\]
\[(3v^2)(2x^2) = 6v^2x^2\]
\[(2x^2)^2 = 4x^4\]
Putting these together, the expression factors as \((3v^2 - 2x^2)(9v^4 + 6v^2x^2 + 4x^4)\).

  1. Complete Factorization over Real Numbers

To determine if the expression can be factored further over the real numbers, we look at the quadratic-like factor \(9v^4 + 6v^2x^2 + 4x^4\). We can rewrite this using the Sophie Germain identity method or by completing the square to express it as a difference of squares:
\[9v^4 + 6v^2x^2 + 4x^4 = (9v^4 + 12v^2x^2 + 4x^4) - 6v^2x^2\]
\[= (3v^2 + 2x^2)^2 - (\sqrt{6}vx)^2\]
This is now a difference of squares, \(A^2 - B^2 = (A - B)(A + B)\), where \(A = 3v^2 + 2x^2\) and \(B = \sqrt{6}vx\). Thus,
\[9v^4 + 6v^2x^2 + 4x^4 = (3v^2 + 2x^2 - \sqrt{6}vx)(3v^2 + 2x^2 + \sqrt{6}vx)\]
Similarly, the factor \(3v^2 - 2x^2\) can be factored as \((\sqrt{3}v - \sqrt{2}x)(\sqrt{3}v + \sqrt{2}x)\). Combining all these, the completely factored form over the real numbers is:
\[(\sqrt{3}v - \sqrt{2}x)(\sqrt{3}v + \sqrt{2}x)(3v^2 + 2x^2 - \sqrt{6}vx)(3v^2 + 2x^2 + \sqrt{6}vx)\]

Examples

Factoring expressions is a foundational skill in algebra that allows engineers and scientists to simplify complex equations. For example, in structural engineering, factoring a stress equation might reveal the critical points (roots) where the material will fail, allowing designers to avoid these regions. Recognizing patterns like the difference of cubes can turn an intimidating polynomial into a manageable product of simpler expressions, making calculations much more efficient.