What is the sum of all solutions to the equation 3x^(2/3) = 54 + 9x^(1/3) A) 63 B) 189 C) 216 D) 243 E) 567 Answer: B Source: GMAT Prep Now Difficulty level: ≈700

Rearrange the equation: 3x(2/3)-9x^(1/3)-54 = 0 Let Y = X(1/3) to simplify things 3Y^2-9Y-54 = 0 >divide through by 3 Y^2-3Y-18=0 Factor: (Y-6)*(y+3) = 0 > Y=6 or Y=-3 X=Y^3, therefore X= 216 or -27 Adding solutions together 216-27=189, B

Brent@GMATPrepNow wrote:What is the sum of all solutions to the equation 3x^(2/3) = 54 + 9x^(1/3) A) 63 B) 189 C) 216 D) 243 E) 567 We should start by recognizing that this is a QUADRATIC EQUATION in disguise. Notice that x^(2/3) = [x^(1/3)]² So let's let x^(1/3) = k, and replace x^(1/3) with k to get: 3k² = 54 + 9k Rearrange to get: 3k² - 9k - 54 = 0 Factor: 3(k² - 3k - 18) = 0 Factor more: 3(k - 6)(k + 3) = 0 So, the solutions are k = 6 and k = -3 Since k = x^(1/3), we can write: x^(1/3) = 6 and x^(1/3) = -3 If x^(1/3) = 6, then x = 6³ = 216 If x^(1/3) = -3, then x = (-3)³ = -27 So, the sum of all solutions = 216 + (-27) = 189 Answer: B Cheers, Brent

I don't think the GMAT would ask this, though, because it can be solved with a simple formula from Algebra II, Viète's Formula. (It isn't really their style to ask questions that have trivial solutions if you know slightly higher math than they expect of you.)

With that in mind, let me propose an alternative solution anyway. Let z equal the cube root of x. With that, we can express the original equation as 3z² - 9z - 54 = 0 z² - 3z - 18 = 0 (z + 3) * (z - 6) = 0 So our solutions are z = -3 and z = 6. z is the cube root of x, so our x's are -3³ and 6³, and their sum = 189.