Question

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

Answer 1

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Answer 2

Answer:

The answer is 1

Step-by-step explanation:

Given:

a + $\frac{1}{b}$ = 3    

b+ $\frac{1}{c}$ = 4  

c+ $\frac{1}{a}$ = $\frac{9}{11}$  

Adding all the given equations we get :

a + b + c + $\frac{1}{a}$ + $\frac{1}{b}$ + $\frac{1}{c}$ = $\frac{86}{11}$

Multipying these equations we get :

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

On further simplifying this equation we get,

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

Putting the value from above equation:

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

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

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

Now taking abc as p,

p+ $\frac{1}{p}$ = $\frac{22}{11}$ =2

Multiplying p on both sides,

$p^{2}$+1 = 2p

$p^{2}$-2p+1 = 0

$p^{2}$-p-p+1 = 0

p(p-1)-1(p-1)=0

p-1=0

p=1

Hence value of a*b*c =1