Question

prove that (a+b+c)² - (a-b-c)² + 4b² - 4c² = (4b+4c)(a+b-c)

please solve this as soon as possible​

Answer 1

Step-by-step explanation:

We have:

(a + b + c)² - (a - b - c)² + 4b² - 4c²

To prove:

(a + b + c)² - (a - b - c)² + 4b² - 4c² = (4b + 4c)(a + b - c)

Solution:

We see:

• (a + b + c)² - (a - b - c)²
• 4b² - 4c²

We know:

• a² - b² = (a + b)(a - b)

Hence:

• (a + b + c + a - b - c)(a + b + c - a + b + c) + 4(b + c)(b - c)
• (2a)(2b + 2c) + 4(b + c)(b - c)
• 4a(b + c) + 4(b + c)(b - c)
• 4(b + c)(a + b - c)
• 4(b + c)(a + b - c)

Therefore:

(a + b + c)² - (a - b - c)² + 4b² - 4c² = (4b + 4c)(a + b - c)

L.H.S = R.H.S