Question

If a+b+c = 6 ,a 2 + b2 + c2 =12 ,find the value of a3 + b3 + c3 -3abc​

Answer 1

Answer :

Let us recall algebraic identity to solve this question

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

which can be written as

a³ + b³ + c³ - 3abc = ( a + b + c){ a² + b² + c² - ( ab + bc + ca) } ----- eq( 1 )

To find the value we need to know the value of ab + bc + ca.

Let's use another algebraic identity to find the value of ab + bc + ca i.e,

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

Substitute the give values in the above identity

6² = 12 + 2( ab + bc + ca)

36 = 12 + 2( ab + bc + ca)

36 - 12 = 2( ab + bc + ca)

24 / 2 = ab + bc + ca

ab + bc + ca = 12

Substitute the value of ab + bc + ca and the give values in eq( 1 )

a³ + b³ + c³ - 3abc = 6( 12 - 12 ) = 6( 0 ) = 0

Therefore the value of a³ + b³ + c³ - 3abc is 0.