6 TST for BMO

Answer: NO. Let us argue by contradiction. Arrange the numbers in $A$ and $B$ in increasing order $1\le a_1<a_2<\cdots <a_{n+1}\le 2^n$ and $2^{n-1}<b_1<b_2\le 2^n$. Apparently, there exists an index $i\le n$ such that $a_i<a_{i+1}\le 2a_i$. We denote: $$a_i{b}_1-1=q_1^3,a_{i+1}{b}_2-1=q_2^3,a_i{b}_2-1=s_1^3,a_{i+1}{b}_1-1=s_2^3.$$ From $a_i<a_{i+1}\le 2a_i$ and $b_1<b_2<2b_1$ it follows $$q_1<s_1,s_2<q_2<2q_1(1).$$

$$(q_1{q}_2{)}^3+q_1^3+q_2^3+1=(s_1{s}_2{)}^3+s_1^3+s_2^3+1(2).$$

Using (1) it can be seen that if $q_1{q}_2>s_1{s}_2$ or $q_1{q}_2<s_1{s}_2$ the equality (2) cannot hold. Thus, $q_1{q}_2=s_1{s}_2$. Let us consider the function $f(x)=x^3+\frac{C}{x^3}$ in the interval $x\in [q_1,q_2]$, where $C=(q_1{q}_2{)}^3$. It decreases in $[q_1,\sqrt{q_1{q}_2}]$ and increases in $[\sqrt{q_1{q}_2},q_2]$, hence it attains its maximum value at the ends of this interval, $f(q_1)=f(q_2)=q_1^3+q_2^3$. Therefore, the equality (2), provided that (1) holds, is true only if $(q_1,q_2)=(s_1,s_2)$, which is the needed contradiction. $\square$