Question

8y3 - 27 = ?
step by step explanation ​

Answer 1

Step-by-step explanation:

${(8)y}^{3} - 27$

$( {2y)}^{ 3} - {(3)}^{3}$

Usind Identity of

$( {a}^{3} - {b}^{3} ) = (a - b)( {a} ^{2} + ab + {b}^{2} )$

It can be written as

$(2y - 3)(4 {y}^{2} + 6y + 9)$

Answer 2

Factor 8y^3-27

8y3−278y3-27

Rewrite 8y38y3 as (2y)3(2y)3.

(2y)3−27(2y)3-27

Rewrite 2727 as 3333.

(2y)3−33(2y)3-33

Since both terms are perfect cubes, factor using the difference of cubes formula, a3−b3=(a−b)(a2+ab+b2)a3-b3=(a-b)(a2+ab+b2) where a=2ya=2y and b=3b=3.

Apply the product rule to 2y2y.

(2y−3)(22y2+2y⋅3+32)(2y-3)(22y2+2y⋅3+32)

Raise 22 to the power of 22.

(2y−3)(4y2+2y⋅3+32)(2y-3)(4y2+2y⋅3+32)

Multiply 33 by 22.

(2y−3)(4y2+6y+32)