mariadono2007 mariadono2007
  • 22-03-2024
  • Mathematics
contestada

216-729x^9 sum/difference of perfect cubes cont

Respuesta :

hudaburhan17 hudaburhan17
  • 22-03-2024

Answer:

(6 - 9x^3)(36 + 54x^3 + 81x^6)

Step-by-step explanation:

To factor the expression 216 - 729x^9 as a sum or difference of perfect cubes, we first need to express each term as a cube.

First, let's rewrite 216 and 729x^9 as cubes:

216 = 6^3

729x^9 = (9x^3)^3

Now, we can use the formula for the difference of cubes:

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Substituting our values:

216 - 729x^9 = (6^3 - (9x^3)^3)(6^2 + 6(9x^3) + (9x^3)^2)

= (6 - 9x^3)(36 + 54x^3 + 81x^6)

So, (216 - 729x^9) can be factored as (6 - 9x^3)(36 + 54x^3 + 81x^6)

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