(x-2y)(x^2+2xy+4y^2)+16y^3 Biết X+2y=0

2 min read Jun 17, 2024

Solving the Expression (x-2y)(x^2+2xy+4y^2)+16y^3 given x+2y=0

This problem involves simplifying an algebraic expression using a given condition. Let's break it down step-by-step:

1. Recognizing the Pattern

Notice that the expression (x^2+2xy+4y^2) resembles the expansion of a cube: (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3

In this case, a = x and b = 2y.

2. Applying the Cube Formula

We can rewrite the expression as:

(x - 2y)(x^2 + 2xy + 4y^2) + 16y^3 = (x - 2y)((x)^3 + (2y)^3) + 16y^3

3. Using the Given Condition

We know x + 2y = 0. Solving for x, we get x = -2y.

4. Substituting and Simplifying

Substituting x = -2y into the expression:

(-2y - 2y)((-2y)^3 + (2y)^3) + 16y^3 = -4y(-8y^3 + 8y^3) + 16y^3 = 0 + 16y^3 = 16y^3

Therefore, the value of the expression (x-2y)(x^2+2xy+4y^2)+16y^3 given x+2y=0 is 16y^3.