Final answer:
To find the value of (a+b)²/ab+(b+c)²/bc+(c+a)²/ca when a+b+c=0, substitute the value of c with -a-b and simplify the expression.
Step-by-step explanation:
To find the value of (a+b)²/ab+(b+c)²/bc+(c+a)²/ca when a+b+c=0, we need to substitute the value of a+b+c in the given expression. Since a+b+c=0, we can replace c with -a-b. Substituting these values, we get:
(a+b)²/ab+(b+(-a-b))²/b(-a-b)+(c+a)²/ca
Simplifying further:
(a+b)²/ab+(-a)²/(-a-b)+(c+a)²/ca
Now, we can simplify each term:
(a+b)² = a²+2ab+b²
(-a)² = a²
(c+a)² = c²+2ac+a²
Substituting these values back into the expression:
(a²+2ab+b²)/ab+a²/(-a-b)+(c²+2ac+a²)/ca
Simplifying the fractions:
a²/ab + 2ab/ab + b²/ab + a²/(-a-b) + c²/ca + 2ac/ca + a²/ca
Cancelling out like terms:
2 + 2 + 2
Therefore, the value of (a+b)²/ab+(b+c)²/bc+(c+a)²/ca is 6.