Question

a^2=b+c, b^2=c+a, c2=a+b then the value of 1÷a+1 + 1÷b+1 + 1÷c+1 equal to​

Answer 1

Step-by-step explanation:

$\huge\underline\mathbb\red{AnSwEr}$

a²=b+c

b²=c+a

c²=a+b

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

1/rootb+c+1 +1/rootc+a+1 +1/roota+b+1

rootc+a+1+rootb+c+1/rootb(rootc+a+1)+c(root6+a+1)

+1(rootc+a+1)+1/roota+b+1

$\frac{ \sqrt{c} + a + 1 + \sqrt{b} + c + 1 }{ \sqrt{bc} + a \sqrt{b} + \sqrt{b} + c \sqrt{c} + ca +c } \times \frac{1}{c + 1}$

solve this u get urs answer