Do It The Smart Way!

Multiply the three equations:

$\left(a+\dfrac{1}{b}\right)\left(b+\dfrac{1}{c}\right)\left(c+\dfrac{1}{a}\right)=\left(\dfrac{7}{3}\right)(4)(1)$

$abc+\dfrac{1}{abc}+a+\dfrac{1}{b}+b+\dfrac{1}{c}+c+\dfrac{1}{a}=\dfrac{28}{3}$

$abc+\dfrac{1}{abc}+\dfrac{7}{3}+4+1=\dfrac{28}{3}$

$abc+\dfrac{1}{abc}=2$

$abc=1$

Solve the first equation for $b$ in terms of $a$ :

$b=\frac{3}{7-3a}$

Solve the last equation for $c$ in terms of $a$ :

$c=\frac{a-1}{a}$

Substitute these expressions for $b$ and $c$ into the second equation:

$\frac{3}{7-3a}+\frac{a}{a-1}=4$

Solving this equation yields $a=\frac{5}{3}$ . Substituting this value into the other equations yields $b=\frac{3}{2}$ and $c=\frac{2}{5}$ .

$abc=\left(\frac{5}{3}\right)\left(\frac{3}{2}\right)\left(\frac{2}{5}\right)=\boxed{1}$ .

I tried a trick method and it worked.

abc=1 if and only if [1/(1+a+1/b)]+[1/(1+b+1/c)]+[1/(1+c+1/a)]=1 (an interesting problem to try as an exercise)

Here [1/(1+a+1/b)]+[1/(1+b+1/c)]+[1/(1+c+1/a)]=1/[1+3/7]+1/[4+1]+1/[1+1]=3/10+1/5+1/2=(3+2+5)/10=1, i.e., abc=1.