It is a problem related to the indices

Question

if a+b+c=0 and c-b=2ab

then, prove that (a²+b²+c²)/abc =2

Answer 1

To prove that (a^2+b^2+c^2)/abc = 2 given a+b+c=0 and c-b=2ab, we can start by using the identity:

(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc

We can substitute the given value of a+b+c=0 into the equation, and simplify it as:

0 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc

Rearranging the terms, we get:

a^2 + b^2 + c^2 = -2ab - 2ac - 2bc

Now we can substitute the given value of c-b=2ab into the equation, and simplify it as:

a^2 + b^2 + (c-b)^2 = -2ab - 2ac - 2bc

Expanding the left side, we get:

a^2 + b^2 + c^2 - 2bc + b^2 = -2ab - 2ac - 2bc

Substituting 2ab for c-b, we get:

a^2 + b^2 + c^2 - 2bc + b^2 = -2ab - 2ac - (c-b)2

Simplifying, we get:

a^2 + 2b^2 + 2ab = 0

Dividing both sides by ab, we get:

(a^2 + 2ab + b^2)/ab = -2

Simplifying the left side, we get:

(a+b)^2/ab = -2

Substituting the given value of a+b=-c, we get:

(-c)^2/ab = -2

Simplifying, we get:

c^2/ab = 2

Finally, we can substitute the value of c from the equation c=-a-b into the expression above:

(a^2+b^2+c^2)/abc = (a^2+b^2+(-a-b)^2)/(ab(-a-b)) = (a^2+b^2+a^2+2ab+b^2)/(-ab(a+b)) = 2(a^2+b^2+2ab)/(-ab(a+b))

Using the identity a+b=-c, we can simplify the expression further as:

2(a^2+b^2+2ab)/(-ab(a+b)) = -2(a^2+b^2+2ab)/(abc)

Substituting the value we obtained for c^2/ab, we get:

-2(a^2+b^2+2ab)/(abc) = -2(c^2/ab)

Therefore, we have proved that (a^2+b^2+c^2)/abc = 2 given a+b+c=0 and c-b=2ab.