if $a$,$b$,$c$ $\in$ $\mathbb{R}$ and if $$a+\frac{1}{b}=\frac{7}{3}$$ $$b+\frac{1}{c}=4$$ $$c+\frac{1}{a}=1$$ Then find the value of $abc$

I multiplied the three equations with $bc$, $ca$ and $ab$ respectively we get

$$abc+c=\frac{7bc}{3}$$ $$abc+a=4ac$$ $$abc+b=ab$$ Again multiplying above three equations with $a$,$b$ and $c$ respectively we get $$a(abc)+ac=\frac{7abc}{3}$$ $$b(abc)+ab=4abc$$ $$c(abc)+bc=abc$$ $\implies$


$$ab=(4-b)abc$$ and $$bc=(1-c)abc$$ Multiplying above all three equations we get

$$a^2b^2c^2=(\frac{7}{3}-a)(4-b)(1-c)a^2b^2c^2$$ $\implies$

$$1=(\frac{7}{3}-a)(4-b)(1-c)$$ But from first set of equations $\frac{7}{3}-a=\frac{1}{b}$ and so on finally we get

$$1=\frac{1}{abc}$$ so $$abc=1$$

I feel this method is little long...can i have any other approach

  • $\begingroup$ Your multiplication of three equations in the end should produce $a^2b^2c^2=(\ldots)a^3b^3c^3$ $\endgroup$ Sep 28 '15 at 6:22
  • $\begingroup$ oh yeah then my approach does not work...thanks $\endgroup$ Sep 28 '15 at 6:24

Hint: Multiplying the first three expressions we get $$abc+\left(a+\dfrac{1}{b}\right)+\left(b+\dfrac{1}{c}\right)+\left(c+\dfrac{1}{a}\right)+\dfrac{1}{abc}=\dfrac{28}{3}.$$


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