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User Yelsayed
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2 Answers

7 votes

Answer:

This may be wrong but: (2x - 5y) ((2x)^2 + 10xy + (5y)^2) over (2x + 5y) (4x^2 + 10xy + 25y^2)

Explanation:

User Patan
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11 votes

Answer: 2x − 5y / 2x + 5y

Explanation:

8x^3 − 125y^3/(2x + 5y)3 ÷ 4x^2 + 10xy + 25y^2/4x^2 + 20xy + 25y^2

To divide by a fraction, multiply by its reciprocal.

8x^3 − 125y^3/(2x + 5y)^3 ⋅ 4x^2 + 20xy + 25y^2/4x^2 + 10xy + 25y^2

Simplify the numerator.

Rewrite 8x^3 as (2x)^3.

(2x)^3 − 125y^3/(2x + 5y)^3 ⋅ 4x^2 + 20xy + 25y^2/4x^2 + 10xy + 25y^2

Rewrite 125y^3 as (5y)^3.

(2x)^3 − (5y)^3/(2x + 5y)^3 ⋅ 4x^2 + 20xy + 25y^2/4x^2 + 10xy + 25y^2

Since both terms are perfect cubes, factor using the difference of cubes formula,

a^3 − b^3 = (a − b) (a^2 + ab + b^2) where a = 2x and b = 5y.

(2x − (5y)) ((2x)^2 + 2x (5y) + (5y)^2)/(2x + 5y)^3 4x + 20xy + 25y^2/ 4x^2 + 10xy + 25y^2

Multiply 5 by −1.

(2x − 5y) ((2x)^2 + 2x (5y) + (5y)^2)/(2x + 5y)^3 ⋅ 4x^2 + 20xy + 25y^2/4x^2 + 10xy + 25y^2

Apply the product rule to 2x.

(2x − 5y) (22x^2 + 2x (5y) + (5y)^2)/(2x + 5y)3 ⋅ 4x^2 + 20xy + 25y^2/4x^2 + 10xy + 25y2

Raise 2 to the power of 2.

(2x − 5y) (4x^2 + 2x (5y) + (5y)^2)/(2x + 5y)^3 ⋅ 4x^2 + 20xy + 25y^2/4x^2 + 10xy + 25y^2

Rewrite using the commutative property of multiplication.

(2x − 5y) (4x^2 + 2 ⋅ 5xy + (5y)^2)/(2x + 5y)^3 ⋅ 4x^2 + 20xy + 25y^2/4x^2 + 10xy + 25y^2

Multiply 2 by 5.

(2x − 5y) (4x^2 + 10xy + (5y)^2)/(2x + 5y)^3 ⋅ 4x^2 + 20xy + 25y^2/ 4x^2 + 10xy + 25y^2

Apply the product rule to 5y.

(2x − 5y) (4x^2 + 10xy + 52y^2)/(2x + 5y)^3 ⋅ 4x^2 + 20xy + 25y^2/4x^2 + 10xy + 25y^2

Raise 5 to the power of 2.

(2x − 5y) (4x^2 + 10xy + 25y^2)/(2x + 5y)^3 ⋅ 4x^2 + 20xy + 25y^2/4x^2 + 10xy + 25y^2

Cancel the common factor of 4x^2 + 10xy + 25y^2.

((2x − 5y)/(2x + 5y)^3) (4x^2 + 20xy + 25y^2)

Multiply

(2x − 5y) (4x^2 + 20xy + 25y^2)/ (2x + 5y)^3

Factor using the perfect square rule.

Rewrite 4x^2 as (2x)^2.

(2x − 5y) ((2x)^2 + 20xy + 25y^2)/ (2x + 5y)^3

Rewrite 25y^2 as (5y)^2.

(2x − 5y) ((2x)^2 + 20xy + (5y)^2)/ (2x + 5y)^3

Check that the middle term is two times the product of the numbers being squared in the first

term and third term.

20xy = 2 ⋅ (2x) ⋅ (5y)

Rewrite the polynomial.

(2x − 5y) ((2x)2 + 2 ⋅ (2x) ⋅ (5y) + (5y)2)/(2x + 5y)^3

Factor using the perfect square trinomial rule a^2 + 2ab + b^2 = (a + b)^2, where a = 2x and b = 5y.

(2x − 5y) (2x + 5y)^2/ (2x + 5y)^3

Cancel the common factor of (2x + 5y)^2 and (2x + 5y)^3.

Factor (2x + 5y)^2 out of (2x − 5y) (2x + 5y)^2.

(2x + 5y)^2 (2x − 5y) / (2x + 5y)^3

Cancel the common factors.

2x − 5y / 2x + 5y

User DotNetBeginner
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