Question

prove that

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

Answer 1

Answer:-

To Prove:-

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

Using (a - b)² = a² + b² - 2ab in RHS we get,

⟹ 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = a² + b² - 2ab + b² + c² - 2bc + c² + a² - 2ca

⟹ 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = a² + a² + b² + b² + c² + c² - 2ab - 2bc - 2ca

⟹ 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 2a² + 2b² + 2c² - 2ab - 2bc - 2ca

⟹ LHS = RHS

Hence, Proved.

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Some important formulae:-

• (a + b)² = a² + b² + 2ab

• a² - b² = (a + b)(a - b)

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

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

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)

• (x + a)(x + b) = x² + (a + b)x + ab

• a² + b² = (a + b)² - 2ab

• a² + b² = (a - b)² + 2ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² = (a - b)² + 4ab

• (a + b)² - (a - b)² = 4ab

Answer 2

Answer:

To Prove :-

$\bullet \sf \: {2a}^{2} + 2 {b}^{2} + 2 {c}^{2} - 2ab - 2bc - 2ca = (a - {b)}^{2} + ( {b - c)}^{2} + {(c - a)}^{2}$

Prove :

Here,

L.H.S = 2a² + 2b² + 2c² - 2ab - 2bc - 2ca

a² + a² + b² + b² + c² + c² - 2ab - 2bc - 2ca

a² - 2ab + b² + b² - 2bc + c² + c² - 2ca + a²

By grouping them

( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( c² - 2ca + a² )

Now,

By uing identity

Using ( a - b )² = a² - 2ab + b²

And putting value

( b - c )² = b² - 2bc + c²

( c - a )² = c² - 2ac + a²

L.H.S = ( a - b )² + ( b - c )² + ( c - a )²

L.H.S = R.H.S

Therefore :-

$\bullet \sf \: {2a}^{2} + 2 {b}^{2} + 2 {c}^{2} - 2ab - 2bc - 2ca = (a - {b)}^{2} + ( {b - c)}^{2} + {(c - a)}^{2}$