Question

If ${\log }_2\left(1+\frac{1}{a}\right)+{\log }_2\left(1+\frac{1}{b}\right)+{\log }_2\left(1+\frac{1}{c}\right)=2$ Find $3abc-(a+b+c+ab+bc+ac)$

Answer 1

$(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})=2^2=4$

$(a+1)(b+1)(c+1)=4abc$

$1+a+b+c+ab+bc+ac=3abc$

$3abc-(a+b+c+ab+bc+ca)=1$