3k

^{3}=27k3k^{3}=27k

Tutors, please sign in to answer this question.

Rome, NY

Hi, Lila!

Vivian and Ken are correct in that two of the roots are +/-3. However, you need to be careful "canceling" out a k on both sides, as this removes one of the possible roots. This type of problem is traditionally solved using factoring:

3k^{3 }= 27k

Rearrange the equation so that it equals zero, just as you would do with a quadratic equation.

3k^{3 }- 27k = 0

Factor out the LCM: 3k

3k(k^{2 }- 9) = 0

Factor the quadratic expression in the parenthesis.

3k(k+3)(k-3) = 0

Set each part equal to 0 and solve for k.

3k = 0 --> k = 0 (this is the root that is removed when you "cancel" out a k early on)

k + 3 = 0 --> k = -3

k - 3 = 0 --> k = 3

This method gives you all 3 roots: k = -3, 0, and 3.

Patty

Pittsfield, MA

Divide both sides by K leaving

3k2=27

Divide both sides by 3...

(3k2)/3=27/3

(3k2)/3=27/3

Resulting in

k2=9

k2=9

Square-root both sides...

k=+/-3

k=+/-3

Middletown, CT

Hi Lila;

3k^{3}=27k

First, you have k^{3} on the left side, and k^{1} on the right side. k^{1} cancels on both sides...

3k^{2}=27

Divide both sides by 3...

(3k^{2})/3=27/3

k^{2}=9

Square-root both sides...

k=+/-3

Evan F.

Mathematics graduate looking to enhance your mathematics ability.

Cambria Heights, NY

4.9
(315 ratings)

John P.

Tutor of math and physics, recent college graduate

Short Hills, NJ

5.0
(21 ratings)

- Algebra 1 4027
- Math 9714
- Algebra 5020
- Math Help 5382
- Algebra Word Problem 2462
- Word Problem 5059
- Algebra 2 Question 524
- Precalculus 1531
- Algebra Help 964
- Geometry 1873

## Comments