Question

Розкладіть на множники 1)a³-a²b-ab+a² 2)x²-2xy+y²-81 3)100-4x²-4xy-y² 4)216-a³ 5)27x³+125y³

Answer 1

1) a³-a²b-ab+a²

(a³+a²)+(-a²b-ab)

a²(a+1)-ab(a+1)

(a²-ab)(a+1)

2) x²-2xy+y²-81

(x-y)²- 9²

(x-y+9)(x-y-9)

3)100-4x²-4xy-y²

10²-(2x-y)²

(10-2x-y)(10+2x-y)

4)216-a³

6³-a³

(6-a)(36+6a+a²)

5)27x³+125y³

(3x)³+(5y)³

(3x+5y)(9x²-15xy+25y²)

^o^ meow <3

Answer 2

1. To factorize a³ - a²b - ab + a², we can factor out the common term 'a²' from the first two terms and the common term 'b' from the last two terms:

a³ - a²b - ab + a² = a²(a - b) - b(a - b) = (a² - b)(a - b).

Therefore, a³ - a²b - ab + a² can be factored as (a² - b)(a - b).

2. The expression x² - 2xy + y² - 81 can be factored as a difference of squares and a difference of squares formula:

x² - 2xy + y² - 81 = (x - y)² - 81 = (x - y + 9)(x - y - 9).

Hence, x² - 2xy + y² - 81 can be factored as (x - y + 9)(x - y - 9).

3. To factorize 100 - 4x² - 4xy - y², we can rewrite it as the square of a binomial:

100 - 4x² - 4xy - y² = (10 - 2x - y)(10 + 2x + y).

Thus, 100 - 4x² - 4xy - y² can be factored as (10 - 2x - y)(10 + 2x + y).

4. To factorize 216 - a³, we can use the difference of cubes formula:

216 - a³ = (6 - a)(36 + 6a + a²).

Therefore, 216 - a³ can be factored as (6 - a)(36 + 6a + a²).

5. The expression 27x³ + 125y³ can be factorized using the sum of cubes formula:

27x³ + 125y³ = (3x)^3 + (5y)^3 = (3x + 5y)(9x² - 15xy + 25y²).

Hence, 27x³ + 125y³ can be factored as (3x + 5y)(9x² - 15xy + 25y²).