# Difference between revisions of "2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 2"

## Problem

If $f(x) = x^3 + 6x^2 + 12x + 6$, solve the equation $f(f(f(x))) = 0.$

## Solution

We can write $f(x) = x^3 + 6x^2 + 12x + 6 = (x+2)^3 -2$. This means that $f(x) = 0 \Rightarrow x = \sqrt[3]{2} - 2$.

Working backwards, $f(f(f(x))) = 0 \Rightarrow f(f(x)) = \sqrt[3]{2} - 2$.

Then, $f(f(x)) = (f(x)-2)^3 - 2 = \sqrt[3]{2} - 2 \Rightarrow f(x) = \sqrt[9]{2}$.

Finally, $f(x) = (x-2)^3 - 2 = \sqrt[9]{2} \Rightarrow x = \sqrt[3]{2+\sqrt[9]{2}}$.

## Solution 2

Note that $f(x)=(x+2)^3-2$. Thus, $f^{n}(x)=(x+2)^{3n}-2$. In the question, $n=3$, so we have $(x+2)^{27}-2=0$. Solving, $\boxed{x=-2+\sqrt[27]{2}}$

~yofro

## See also

 2014 UNM-PNM Contest II (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 All UNM-PNM Problems and Solutions
Invalid username
Login to AoPS