Question

a+b+c=22 and ab+bc+ca=91abc then the valu of a(b*b+c*c)+b(c*c+a*a)+c(a*a+b*b)/abc

Answer 1

Answer:

1999

I hope this helps you.

Step-by-step explanation:

Use

a(b²+c²) + b(c²+a²) + c(a²+b²) = (a+b+c) (ab+bc+ca) - 3abc

How to discover this?  We're given a+b+c, which has terms of degree 1, and we're given some information about ab+bc+ca, which has terms of degree 2.  Since a(b²+c²) + etc. has degree 3, it looks like it might be good to try multiplying a+b+c by ab+bc+ca.  Turns out to be a good move!

So

[ a(b²+c²) + b(c²+a²) + c(a²+b²) ] / abc

= (a+b+c) × (ab+bc+ca) / abc  -  3

= 22 × 91  -  3

= 2002 - 3

= 1999