Can anyone factor 27a^4 - a ?

Question

Explanation: 27a to the 4th exponent minus a.

Answer 1

27a^4 - a
= a(27a^3 - 1)
= a[(3^3)(a^3) - 1]
= a [(3a)^3 - 1]
= a (3a - 1)[(3a)^2 + (3a)(1) + 1^2]
= a (3a - 1) (9a^2 + 3a + 1)

Note:
Difference of Cubes formula:
a^3 - b^3
= (a - b)(a^2 + ab + b^2)

Answer 2

Factor out a to get a* (27a^3-1). The term in ( )
has a difference of cubes solution, which I believe is (3a-1)(9a^2+9a+1). You might want to check this form in your book.

Answer 3

OK, I will do this one eventhough yu got so many answers:

facor out an a to get:

a (27a³ - 1) =

now you have the diff of cubes:

a(3a -1)(9a² + 3a + 1)

Answer 4

1) look for common factors.

note you can factor out a, as
a(27a^3 - 1).

2) Look for special products, like A^3 - B^3. Note this is

(A-B)(A^2 + AB + B^2), so

a(27a^3 -1) = a( [3a]^3 - [1]^3), or A = 3a, B = 1 in formula
= a(3a-1)( [3a]^2 + [3a][1] + [1]^2) or
= a(3a-1)(9a^2 + 3a + 1)

Answer 5

27a^4-a
a(27a^3-1)
a(3a-1)(.......? )
(....?...) is (9a^2+3a+1)
I got that by dividing (3a-1) into (27a^3-1). I was careful to write this as (27a^3+0a^2+0a-1) /(3a-1)
I don't know how you like to do this kind of division, so I leave this detail to you. However you do it, your answer should be (9a^2+3a+1), the same as mine.

The final answer is 27a^4-a =a(3a-1)(9a^2+3a+1)

Answer 6

27a^4 - a=a(27a^3-1)
=a(3a-1)(9a^2+3a+1)

Answer 7

27a^4-a
=a[27a^3-1]
=a[(3a)^3- 1^3]
=a(3a-1)[9a^2+3a+1]. ANS.

Answer 8

take out one "a"

a(27a^3-1)

Answer 9

a*(3*a-1)*(9*a^2+3*a+1)

Answer 10

Go to www.webmath.com they have a great little section that does the factor and shows you the steps