How do you factor $x^{3} - 81$ $x^3 - 81$ ?

Question

How do you factor $x^{3} - 81$ $x^3 - 81$ ?

1 Answer

Impact of this question

9804 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-10T11:35:06

Express as a difference of cubes, then use the difference of cubes identity to find:

$x^{3} - 81 = (x - 3^{\frac{4}{3}}) (x^{2} + 3^{\frac{4}{3}} x + 3^{\frac{8}{3}})$ $x^3-81 = (x-3^(4/3))(x^2+3^(4/3)x+3^(8/3))$

Explanation:

When $a \geq 0$ $a >= 0$ and $b, c \neq 0$ $b, c != 0$

$a^{b c} = {(a^{b})}^{c}$ $a^(bc) = (a^b)^c$

So:

$81 = 3^{4} = 3^{\frac{4}{3} \cdot 3} = {(3^{\frac{4}{3}})}^{3}$ $81 = 3^4 = 3^(4/3 * 3) = (3^(4/3))^3$

The difference of cubes identity is:

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

So:

$x^{3} - 81 = x^{3} - {(3^{\frac{4}{3}})}^{3}$ $x^3 - 81 = x^3 - (3^(4/3))^3$

$= (x - 3^{\frac{4}{3}}) (x^{2} + x (3^{\frac{4}{3}}) + {(3^{\frac{4}{3}})}^{2})$ $=(x-3^(4/3))(x^2 + x(3^(4/3)) + (3^(4/3))^2)$

$= (x - 3^{\frac{4}{3}}) (x^{2} + 3^{\frac{4}{3}} x + 3^{\frac{8}{3}})$ $=(x-3^(4/3))(x^2+3^(4/3)x+3^(8/3))$

Science

Math

Humanities

... and beyond

How do you factor $x^{3} - 81$ $x^3 - 81$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you factor x^3 - 81x3−81x^3 - 81?

1 Answer

Explanation:

Related questions

Impact of this question

How do you factor $x^{3} - 81$ $x^3 - 81$ ?