QuestionJuly 15, 2025

Factor using the sum or difference of cubes. 4x^6+32y^3= square

Factor using the sum or difference of cubes. 4x^6+32y^3= square
Factor using the sum or difference of cubes. 4x^6+32y^3= square

Solution
4.7(160 votes)

Answer

(2x^2 + 2y)(4x^4 - 4x^2y + 4y^2) Explanation 1. Identify the sum or difference of cubes Recognize 4x^6 + 32y^3 as a sum of cubes. Rewrite it as (2x^2)^3 + (2y)^3. 2. Apply the sum of cubes formula Use the formula a^3 + b^3 = (a + b)(a^2 - ab + b^2) where a = 2x^2 and b = 2y. 3. Substitute and simplify Substitute a and b: (2x^2 + 2y)((2x^2)^2 - (2x^2)(2y) + (2y)^2). Simplify: (2x^2 + 2y)(4x^4 - 4x^2y + 4y^2).

Explanation

1. Identify the sum or difference of cubes<br /> Recognize $4x^6 + 32y^3$ as a sum of cubes. Rewrite it as $(2x^2)^3 + (2y)^3$.<br /><br />2. Apply the sum of cubes formula<br /> Use the formula $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ where $a = 2x^2$ and $b = 2y$.<br /><br />3. Substitute and simplify<br /> Substitute $a$ and $b$: $(2x^2 + 2y)((2x^2)^2 - (2x^2)(2y) + (2y)^2)$.<br /> Simplify: $(2x^2 + 2y)(4x^4 - 4x^2y + 4y^2)$.
Click to rate:

Similar Questions