"Factor each of the algebraic expressions completely. 27x3−8y3" was answered by our community

High School
Algebra
Algebra I
Algebra foundations

bmgf3m

Answered question

2021-12-17

Factor each of the algebraic expressions completely.

27 x^{3} - 8 y^{3}

Answer & Explanation

Samantha Brown

Beginner2021-12-18Added 35 answers

27 x^{3} - 8 y^{3}

The formula for this expression is:

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

where

a = 3 x and b = 2 y

{(3 x)}^{3} - {(2 y)}^{3} = (3 x - 2 y) ({(3 x)}^{2} + (3 x) \times (2 y) + {(2 y)}^{2})

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

Joseph Lewis

Beginner2021-12-19Added 43 answers

27 x^{3} - 8 y^{3}

= {(3 x)}^{3} - {(2 y)}^{3}

= (3 x - 2 y) ({(3 x)}^{2} + (3 x) (2 y) + {(2 y)}^{2})

= (3 x - 2 y) (3^{2} x^{2} + 6 x y + 2^{2} y^{2})

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

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

This is factored all the way.

RizerMix

Expert2021-12-29Added 656 answers

The formula for factoring the difference of cubes is: $u^{3} - v^{3} = (u - v) \cdot (u^{2} + u v + v^{2})$
$27 x^{3} - 8 y^{3} = (3 x)^{3} - (2 y)^{3}$
$u = 3 x a n d v = 2 y$
$(3 x - 2 y) \cdot ((3 x)^{2} + (3 x) \cdot (2 y) + (2 y)^{2})$
$(3 x - 2 y) \cdot (9 x^{2} + 6 x y + 4 y^{2})$

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