Question

If a – b + c = 4, ab + bc – ac = 17, find the value of a² + b² + c².

Answer 1

Given: a – b + c = 4, ab + bc – ac = 17

Step-wise calculation:

1. Start with the expression (a – b + c)²: 

(a – b + c)² = a² + b² + c² – 2ab + 2bc – 2ac

2. Substitute the given values:

(a – b + c)² = 4²

(a – b + c)² = 16

a² + b² + c² – 2(ab – bc + ac) = 16

3. Notice the given second equation (ab + bc – ac = 17).

Rearranging it to fit the middle terms: ab + bc – ac = 17.

So, –2(ab – bc + ac) = –2(ab – bc + ac).

We need carefully to rewrite middle terms:

a² + b² + c² – 2ab + 2bc – 2ac = 16

Observe: 2ab + 2bc – 2ac = –2(ab – bc + ac)

If we let S = ab – bc + ac, then a² + b² + c² – 2S = 16.

We are given: ab + bc – ac = 17.

But we need (ab – bc + ac) to substitute for (S).

4. Find (S = ab – bc + ac) in terms of the given expression.

Note: (ab + bc – ac) + (ab – bc + ac) = 2ab.

So, 17 + S = 2ab

⇒ S = 2ab – 17.

But we don’t know (ab) directly.

5. Alternatively, re-express (S) in terms of known quantities: 

Compare (S) and the given S = ab – bc + ac 

Given: T = ab + bc – ac = 17

Add (S + T): S + T = (ab – bc + ac) + (ab + bc – ac)

S + T = 2ab 

S + 17 = 2ab

⇒ S = 2ab – 17

6. Let’s use the identity: 

(a – b + c)² = a² + b² + c² – 2ab + 2bc – 2ac 

We were given: (a – b + c)² = 16 

Substitute: a² + b² + c² – 2(ab – bc + ac) = 16

Rewrite as: a² + b² + c² – 2S = 16

7. Since S = 2ab – 17, 

Substitute back: a² + b² + c² – 2(2ab – 17) = 16

a² + b² + c² – 4ab + 34 = 16 

a² + b² + c² – 4ab = 16 – 34

a² + b² + c² – 4ab = –18

8. We need an expression to find (ab) or (a² + b² + c²).

For this, note the identity: (a – b)² = a² + b² – 2ab

We can try to express (a² + b² + c²) in terms of (ab), but lacking (a + b + c) or (a – b + c) sums, solve for (a² + b² + c²):

Rewrite: a² + b² + c² – 4ab = –18

Add and subtract (2bc – 2ac):

a² + b² + c² + 2bc – 2ac – 4ab – 2bc + 2ac = –18

Group terms: (a² + b² + c² + 2bc – 2ac) – 4ab – 2bc + 2ac = –18 

Recall from step 1: a² + b² + c² – 2ab + 2bc – 2ac = 16 

Then, 16 – 2ab = –18 

–2ab = –34 

ab = 17

9. Substitute (ab = 17) back into the earlier equation: 

a² + b² + c² – 4(17) = –18

a² + b² + c² – 68 = –18 

a² + b² + c² = 50