Question

If (a^2+b^2+c^2)=89 and (ab-bc-ca)=16 find the value of (a+b-c)

Answer 1

Answer:

We know that

(a+b-c)² = a²+b²+c² + 2 (ab-bc-ca)

substituting given values

(a+b-c)² = 89+2(16)

(a+b-c)² = 121

(a+b-c) = 11

HOPE IT IS HELPFUL

Answer 2

Given:

• (a² + b² + c²) = 89
• ab - bc - ca = 16

To Find:

• a + b - c

Solution:

We know that;

⇒ (a + b + (-c))² = a² + b² + (-c)² + 2ab + 2(b)(-c) + 2(-c)(a)

(Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca)

⇒ (a + b + (-c))² = 89 + 2(ab - bc -ca)

⇒ (a + b - c)² = 89 + 2(16)

⇒ (a + b - c)² = 89 + 32

⇒ (a + b - c)² = 121

⇒ (a + b - c) = √121

⇒ (a + b - c) = 11

∴ Final Answer: (a + b - c) = 11