Question

if a+b+c = 3*(3^1/2) and a^2 + b^2 + c^2 = 27, prove that ab+bc+ca = 0​

Answer 1

Solution :

Given -

a + b + c = 3 × 3^½

& a² + b² + c² = 27

To prove :

ab+bc+ca = 0

Proof -

a + b + c = 3 × 3^½

Squaring this on both sides ;

=> [ a + b + c ]² = [ 3 × 3^½ ]²

=> [ a + b + c ]² = [ 3]² × [ 3^½ ]²

=> [ a + b + c ]² = 9 × 3

=> [ a + b + c ]² = 27

=> a² + b² + c² + 2 ( ab + bc + ca ) = 27

But we know that a² + b² + c² = 27

=> 27 + 2( ab + bc + ca ) = 27

=> 2( ab + bc + ca ) = 27 - 27 = 0

=> ab + bc + ca = 0 .

Hence Proved

_______________________________

Additional Information :

(a + b)² = a² + 2ab + b²

(a + b)² = (a - b)² + 4ab

(a - b)² = a² - 2ab + b²

(a - b)² = (a + b)² - 4ab

a² + b² = (a + b)² - 2ab

a² + b² = (a - b)² + 2ab

2 (a² + b²) = (a + b)² + (a - b)²

4ab = (a + b)² - (a - b)²

ab = {(a + b)/2}² - {(a-b)/2}²

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

(a + b)³ = a³ + 3a²b + 3ab² b³

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)( a² - ab + b² )

a³ + b³ = (a + b)³ - 3ab( a + b)

a³ - b³ = (a - b)( a² + ab + b²)

a³ - b³ = (a - b)³ + 3ab ( a - b )

_______________________________

Answer 2

Given

• A + B + C = 3 × 3^½

• A² + B² + C² = 27

We Find

Prove :- AB + BC + CA = 0

According to the question

By Squaring Both Sides,

$\implies \sf {[ A + B + C ]² = [ 3 × 3^½ ]² }$

$\implies \sf {[ A + B + C ]² = [ 3 ]² × [ 3^½ ]² }$

$\implies \sf {[ A + B + C ]² = 9 × 3 }$

$\implies \sf {[ A + B + C ]² = 27 }$

$\implies \sf { A² + B² + C² + 2(AB + BC + CA) = 27 }$

$\implies \sf { We\:Know,\: A² + B² + C² = 27}$

$\implies \sf { 27 + 2(AB + BC + CA) = 27 }$

$\implies \sf { 2(AB + BC + CA) = 27 - 27 = 0}$

$\implies \sf { AB + BC + CA = 0}$

Hence, We prove that AB + BC + CA = 0.