Question

Let $X=A\overline{BC}+B\overline{CA}+C\overline{AB}$.Evaluate $X$ for
1.$A=1,B=0,C=1$
2.$A=B=C=1$ and
3.$A=B=C=0$

Answer 1

$Part1)A=1,B=0,C=1$

It is binary operation. substitute the value

$X=1.\overline{\left(0.1\right)}+0.\overline{\left(1.1\right)}+1.\overline{\left(1.0\right)}$

$X=1.\overline{0}+0.\overline{1}+1.\overline{0}$ $\because 1.1=0,1.0=0$

$X=1.1+0.0+1.1$ $\because 0=1,1=0$

$X=1+0+1$ $\because 1+1=1$

$X=1+1=1$

$part2)A=B=C=1$

$X=1.\overline{\left(1.1\right)}+1.\overline{(1.1})+1.\overline{\left(1.1\right)}$

$X=1.\overline{1}+1.\overline{1}+1.\overline{1}$

$X=1.0+1.0+1.0$

$X=0+0+0=0$

$Part3)A=B=C=0$

$X=0.\overline{\left(0.0\right)}+0.\overline{(0.0})+0.\overline{\left(0.0\right)}$

$X=0.\overline{0}+0.\overline{0}+0.\overline{0}$

$X=0.1+0.1+0.1$

$X=0+0+0=0$