Question

If a+b+c=0 then find a^2/bc+b^2/ca+c^2/ab

Answer 1

Answer:

3

Step-by-step explanation:

Given---> a + b + c = 0

To find ---> a + b + c = 0

Solution---> We have an identity

If x + y + z = 0 , then

x³ + y³ + z³ = 3xyz , applying it here , we get

given , a + b + c = 0

So , a³ + b³ + C³ = 3 abc

Now , returning to original problem,

a²/bc + b² / ca + c² / ab

Taking abc , LCM , we get

= (a³ + b³ + c³ ) / abc

Putting , a³ + b³ + c³ = 3abc in it

= 3 abc / abc

= 3

Additional identities--->

(1) (a + b )² = a² + b² + 2ab

(2) ( a- b )² = a² + b² - 2ab

(3) a² - b² = ( a + b ) ( a - b )

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

(5) ( a + b )³ = a³ + b³ + 3ab ( a + b )

(6) ( a - b )³ = a³ - b³ - 3ab ( a - b )

Answer 2

Question :------ If a+b+c=0 , then, find a^2/bc+b^2/ca+c^2/ab ......

Formula to be used

• if a+b+c = 0 , a³+b³+c³ = 3abc

$\frac{ {a}^{2} }{bc} + \frac{ {b}^{2} }{ca} + \frac{ {c}^{2} }{ab} \\ \\ \implies \: \frac{ {a}^{3} + {b}^{3} + {c}^{3} }{abc} \\ \\ \implies \: \frac{3abc}{abc} \\ \\ \implies \: 3$

Additional brainly knowledge :----

[1] ( a + b )² = a² + 2ab + b²

[2] ( a – b )² = a² – 2ab + b²

[3] ( a + b )³ = a³ + 3a² b + 3ab² + b³

[4] ( a – b )³ = a ³ – 3a² b + 3ab² – b³

[5] ( a + b )( a ² - ab + b² ) = a³ + b³

[6] ( a – b )( a ² + ab + b² ) = a³ – b³