Question

if a+1/b=3, b+1/c=4,c+1/a=9/11 then find a×b×c

Answer 1

Answer:

Given an equation: $a+\frac{1}{b}=3$ , $b+\frac{1}{c}=4$ and $c+\frac{1}{a} =\frac{9}{11}$.

Sum of these equation is, $a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=7+\frac{9}{11} =\frac{86}{11}$.

Product of the above equation:  $(a+\frac{1}{b})\cdot (b+\frac{1}{c})\cdot (c+\frac{1}{a})$

On Simplify:

$abc+a+c+\frac{1}{b} +b+\frac{1}{c}+\frac{1}{a} +\frac{1}{abc} = \frac{108}{11}$

or

abc+$(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$+$\frac{1}{abc}$= $\frac{108}{11}$.

$abc+\frac{86}{11}+\frac{1}{abc}=\frac{108}{11}$

$abc+\frac{1}{abc} = \frac{108}{11}-\frac{86}{11} = \frac{108-86}{11} =\frac{22}{11}$

⇒ $abc+\frac{1}{abc} =2$.

Let abc be x

then;  $x+\frac{1}{x} = 2$ or

$x^2-2x+1=0$

On solving the quadratic equation; we have,

x=1 or

$a\times b \times c=1$

Answer 2

Answer:

Step-by-step explanation:

Answer:

Given an equation:  ,  and .

Sum of these equation is, .

Product of the above equation:  

On Simplify:

or

abc++= .

⇒ .

Let abc be x

then;   or

On solving the quadratic equation; we have,

x=1 or