2010/064) if a+b+c=1, ab+bc+ca=2, abc=3, then

find the value of 1/(a+bc)+1/(b+ca)+1/(c+ab).


we have

a+b+c=1 ...1

ab+bc+ca=2 .. 2

abc=3 ...3

so a b c are roots of equation

x^3-x^2 + 2x - 3 = 0

further

as a+ b+ c = 1

so a = 1- (b+c)
so a + bc = 1-(b+c) + bc = (1-b)(1-c)

similarly b+ ca = (1-c)(1-a)
c + ba = (1-b)(1-a)

we need to find 1/(a+bc)+1/(b+ca)+1/(c+ab) = 1/ ((1-b)(1-c)) + 1/ ((1-c)(1-a)) + 1/ ((1-b)(1-a))

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



so we need to form an equation whose roots are a-1 , b- 1 and c- 1

as a b and c are roots of f(x) = x^3-x^2 + 2x - 3

so a-1 b- 1 and c- 1 are roots of

f(x+1) = (x+1)^3 - (x+1)^2 + 2(x+1) - 3 = x^3 + 2 x^2 + 3x -1

so (a-1) + (b-1) + (c-1) = - 2

(a-1)(b-1)(c-1) = 1

so given expression = ((a-1) + (b-1) + (c-1))/((a-1)(b-1)(c-1)) = - 2