Question

If ab+bc+ac=0, find the value of (1/a²-bc)+(1/b²-ca)+(1/c²-ab)

Answer 1

The value of
`\((1)/(a^2 - bc) + (1)/(b^2 - ca) + (1)/(c^2 - ab)\)` is
`\(0\)`.

How did we get the value?

To find the value of
`\((1)/(a^2 - bc) + (1)/(b^2 - ca) + (1)/(c^2 - ab)\) given \(ab + bc + ac = 0\)`, let's try to express each denominator in terms of the given condition.

Given
`\(ab + bc + ac = 0\)`, we can rewrite each term in the denominators as follows:

1.
`\(a^2 - bc = a^2 + ac + ab - 2ac - ab = (a + b)(a - c)\)`

2.
`\(b^2 - ca = b^2 + ab + bc - 2bc - ab = (b + c)(b - a)\)`

3.
`\(c^2 - ab = c^2 + ac + bc - 2ac - bc = (c + a)(c - b)\)`

Now, substitute these expressions back into the given expression:

`\[(1)/((a + b)(a - c)) + (1)/((b + c)(b - a)) + (1)/((c + a)(c - b))\]`

Next, factor out the common denominators:

`\[((b + c) + (a - c))/((a + b)(a - c)) + ((c + a) + (b - a))/((b + c)(b - a)) + ((a + b) + (c - b))/((c + a)(c - b))\]`

Combine the numerators:

`\[(2a)/((a + b)(a - c)) + (2b)/((b + c)(b - a)) + (2c)/((c + a)(c - b))\]`

Now, substitute the expressions we found for the denominators:

`\[(2a)/((a + b)(a - c)) + (2b)/((b + c)(b - a)) + (2c)/((c + a)(c - b))\]`

Now, use the given condition
`\(ab + bc + ac = 0\)` to simplify the expression:

`\[(2a)/((a + b)(a - c)) + (2b)/((b + c)(b - a)) + (2c)/((c + a)(c - b))\]`

`\[= (2a)/((-c)(a - c)) + (2b)/((-a)(b - a)) + (2c)/((-b)(c - b))\]`

`\[= (2)/(c) + (2)/(a) + (2)/(b)\]`

Combine the fractions:

`\[= (2(a + b + c))/(abc)\]`

Since
`\(ab + bc + ac = 0\), \(a + b + c = 0\)`, so the expression simplifies to:

`\[= (2 \cdot 0)/(abc) \\ = 0\]`

Therefore, the value of
`\((1)/(a^2 - bc) + (1)/(b^2 - ca) + (1)/(c^2 - ab)\)` is
`\(0\)`.