^{3}=27k

^{3 }= 27k

^{3 }- 27k = 0

^{2 }- 9) = 0

3k^{3}=27k

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Patricia S. | Math Tutoring for K-12 & CollegeMath Tutoring for K-12 & College

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

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

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

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