Question

a+b+c=6,ab+bc+ca=11,thena^3+b^3+c^3-3abc=?

Answer 1

Hi friend, Harish here.

Here is your answer:

Given that,

a + b + c = 6.  & ab + bc + ca = 11

To find,

a³ + b³ + c³ - 3abc .

Solution,

We know that,

$( a + b + c) = 6$      [Square on both sides]

Then, 

$(a + b + c )^{2} = 6^{2} = 36$

⇒ $a^{2} + b^{2}+c^{2} + 2(ab +bc+ca) = 36$

[Now substitute the value of ab+bc +ca]

Then,

⇒ $a^{2} + b^{2}+c^{2} + 2(11) = 36$

⇒ $a^{2} + b^{2}+c^{2} = 36 - 22 = 14$

We know that,

$a^{3}+b^{3}+c^{3}-3abc= (a+b+c)(a^{2}+b^{2}+c^{2} - (ab+bc+ca))$

⇒ $a^{3}+b^{3}+c^{3}-3abc= (6)(14 -11) = 6 \times 3 = 18.$


Therefore the value is 18.
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Hope my answer is helpful to you

Answer 2

Hello friends!! Cutiepie Here..

Here is ur answer :

Given :

a + b + c = 6

ab + bc + ca = 11

To find :

a³ + b³ + c³ - 3abc

Solution :

Using identity : a³ + b³ + c³ - 3abc = ( a + b + c ) ( a² + b² + c² - ab - bc - ca)


a³ + b³ + c³ - 3abc = ( a + b + c ) [a² + b² + c² - (ab + bc + ca)]



We haven't know the value of a² + b² + c²

So, we have to find it.

Identity :

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

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

Now, put all the values.

( 6 )² = a² + b² + c² + 2 ( 11 )

36 = a² + b² + c² + 22

a² + b² + c² = 36 - 22

a² + b² + c² = 14

Now, we have all the values used in eqⁿ (1)

That is, a + b + c

a² + b² + c²

ab + bc + ca

So. Put these values in eqⁿ (1)

a³ + b³ + c³ - 3abc

= ( 6 ) ( 14 - 11 )


= ( 6 ) ( 3 )

= 18

Answer is 18.


HOPE IT HELPS YOU..