Question

(x+y+z)^2 + (x+y+z)^2

Answer 1

Answer:

2(x+y+z)²

Step-by-step explanation:

(x+y+z)²+(x+y+z)²

2(x+y+z)²

Answer 2

Given :

→ (x + y + z)² + (x + y + z)²

Identity to use :

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

Solution :

→ (x + y + z)² + (x + y + z)²

where,

• a = x
• b = y
• c = z

Method 1 :

(x + y + z)² + (x + y + z)²

= 2(x + y + z)²

Applying the identity,

= 2[(x)² + (y)² + (z)² + 2(x)(y) + 2(y)(z) + 2(z)(x)]

= 2[x² + y² + z² + 2xy + 2yz + 2zx]

→ 2x² + 2y² + 2z² + 4xy + 4yz + 4zx

Method 2 :

(x + y + z)² + (x + y + z)²

Applying identity,

= [(x)² + (y)² + (z)² + 2(x)(y) + 2(y)(z) + 2(z)(x)] + [(x)² + (y)² + (z)² + 2(x)(y) + 2(y)(z) + 2(z)(x)]

= (x² + y² + z² + 2xy + 2yz + 2zx) + (x² + y² + z² + 2xy + 2yz + 2zx

= x² + y² + z² + 2xy + 2yz + 2zx + x² + y² + z² + 2xy + 2yz + 2zx

→ 2x² + 2y² + 2z² + 4xy + 4yz + 4zx

IDENTITIES :

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• (x + a)(x + b) = x² + x(a + b) + ab
• a² - b² = (a + b)(a - b)
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• (a + b)³ = a³ + 3a²b + 3ab² + b³ → a³ + 3ab(a + b) + b³
• (a - b)³ = a³ - 3a²b + 3ab² - b³ → a³ - 3ab(a - b) + b³
• a³ + b³ = (a + b)(a² - ab + b²)
• a³ - b³ = (a - b)(a² + ab + b²)
• a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)
• a³ + b³ + c³ = 3abc if a + b + c = 0