Find x cubed with half adder
Let the LSB of the number be \$a\$ and MSB be \$b\$. Then the number is \$2b+a\$ where, \$a,b\in\{0,1\}\$.

$$(2b+a)^3=8b^3 + 12ab^2+6a^2b+a^3$$

Since \$a,b\in\{0,1\}, \ a^3=a^2=a\$ , \$b^3=b^2=b\$ and \$a\times b = a\ AND\ b = ab\$.

$$\therefore (2b+a)^3=8b+18ab+a$$
$$=16ab+8b+2ab+a$$
$$=2^4\times ab+2^3\times b+2^2\times 0+2^1\times ab+2^0\times a$$

From this we can say that,


the \$0^{th}\$ bit (LSB) of answer is \$a\$  
\$1^{st}\$ and \$4^{th}\$ bits are \$ab\$ 
\$2^{nd}\$ bit is \$0\$
\$3^{rd}\$ bit is \$b\$


The 'carry out' of an half adder, with inputs \$a\$ and \$b\$, will be \$ab\$. So circuit can be implemented as given below.



\$\mathtt{ba}\$ is the input and \$\mathtt{Y_4Y_3Y_2Y_1Y_0}\$ is the output (\$\mathtt{Y_0}\$ is the LSB).