\( ( x + 2 y ) ( x - 2 y ) ( x ^ { 2 } - 2 x y + 4 y ^ { 2 } ) ( x ^ { 2 } + 2 x y + \Delta y ^ { 2 } ) \)

The expression you've provided is: \[ (x + 2y)(x - 2y)(x^2 - 2xy + 4y^2)(x^2 + 2xy + \Delta y^2) \] To determine the value of \(\Delta\) that allows this expression to factor nicely, let's analyze the components. ### Understanding the Structure 1. **First Two Factors: \((x + 2y)(x - 2y)\)** This is a difference of squares: \[ (x + 2y)(x - 2y) = x^2 - (2y)^2 = x^2 - 4y^2 \] 2. **Next Two Factors: \((x^2 - 2xy + 4y^2)(x^2 + 2xy + \Delta y^2)\)** Observe that: - The first part, \(x^2 - 2xy + 4y^2\), resembles the expansion of \((x - y)^2 + 3y^2\), which is part of factoring higher-degree polynomials. - The second part, \(x^2 + 2xy + \Delta y^2\), needs to be determined so that the product can be factored into a recognizable form. ### Connection to Perfect Cubes The given expression resembles the factorization of the difference of cubes. Specifically: \[ x^6 - 64y^6 = (x^3)^2 - (8y^3)^2 = (x^3 - 8y^3)(x^3 + 8y^3) \] Each of these can be further factored: \[ x^3 - 8y^3 = (x - 2y)(x^2 + 2xy + 4y^2) \] \[ x^3 + 8y^3 = (x + 2y)(x^2 - 2xy + 4y^2) \] ### Matching the Given Expression Comparing the factorization with your expression: \[ (x + 2y)(x - 2y)(x^2 - 2xy + 4y^2)(x^2 + 2xy + \Delta y^2) \] To align with the standard factorization of \(x^6 - 64y^6\), the term \(x^2 + 2xy + \Delta y^2\) should be: \[ x^2 + 2xy + 4y^2 \] Thus, \(\Delta = 4\). ### Final Value \[ \Delta = 4 \] ### Verification Plugging \(\Delta = 4\) back into the expression: \[ (x + 2y)(x - 2y)(x^2 - 2xy + 4y^2)(x^2 + 2xy + 4y^2) = x^6 - 64y^6 \] This confirms that \(\Delta = 4\) is the correct value for the expression to represent \(x^6 - 64y^6\).