Question

a²/b+c + b²/a+c + c²/a+b=​

Answer 1

Answer:

a² = b + c

b² = c + a

c² = a + b

a² - b² = b - a      =>  (a - b) (a + b) = b - a

               =>  a + b  = -1    or  a = b

1)  a + b ≠ - 1      as  c²  is not negative, if c is a real number.

         a = b  =>  c² = 2 a

     similarly, b = c    =>  a² = 2 b            (as  b+c ≠ -1, as a² is non negative)

       and  c = a =>  b² = 2 a

   so we get a = b = c = 2

\begin{gathered}\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\\\\=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\end{gathered}1+a1+1+b1+1+c1=31+31+31=1

Answer 2

Step-by-step explanation:

In this question

a^2=b+c , b^2=c+a , c^2=a+b is given

Now we are asked to find the value of

1/1+a + 1/1+b + 1/1+c

So to solve this we need the value of 1+a , 1+b , 1+c and to find it what we will do is add a , b and c to the given equations……. okk let’s do it

a^2=b+c

a^2+a = a+b+c

a (a+1)=a+b+c

a+1= (a+b+c)/a

For 1+b also follow same process adding b on both sides and we get

1+b= (a+b+c)/ b

and for 1+c , adding c on both sides , we get

1+c=(a+b+c)/c

Now put these values in the question whose value is to be found

1/1+a + 1/1+b + 1/1+c

a/(a+b+c) + b/(a+b+c) + c/(a+b+c)

Taking LCM :

(a+b+c )/ (a+b+c ) =1

And hence we have solved it , these questions require the need of tricks as we cannot solve it as a whole as it would be quite lengthy process hence we added a ,b and c in the beginning only so that we get something uniform out of it.

All the best ……!

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