# 1991 AHSME Problems/Problem 20

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## Problem

The sum of all real $x$ such that $(2^x-4)^3+(4^x-2)^3=(4^x+2^x-6)^3$ is

$\textbf{(A) } \frac32 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } \frac52 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } \frac72$

## Solution

Note that $(2^x-4)+(4^x-2)=4^x+2^x-6,$ so we let $a=2^x-4$ and $b=4^x-2.$ The original equation becomes $$a^3+b^3=(a+b)^3.$$ We expand the right side, then rearrange: \begin{align*} a^3+b^3 &= a^3+3a^2b+3ab^2+b^3 \\ 0 &= 3a^2b+3ab^2 \\ 0 &= 3ab(a+b). \end{align*}

• If $a=0,$ then $2^x-4=0,$ from which $x=2.$
• If $b=0,$ then $4^x-2=0,$ from which $x=\frac12.$
• If $a+b=0,$ then $4^x+2^x-6=0.$ As $4^x=(2^x)^2,$ we rewrite this equation, then factor: \begin{align*} (2^x)^2+2^x-6&=0 \\ (2^x-2)(2^x+3)&=0. \end{align*} If $2^x-2=0,$ then $x=1.$

If $2^x+3=0,$ then there are no real solutions for $x,$ as $2^x+3>3$ holds for all real numbers $x.$

Together, the answer is $2+\frac12+1=\boxed{\textbf{(E) } \frac72}.$

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