If $$a+b+c=9$$ and $$ab+bc+ca=-22$$, then the value of  $$a^{3}+b^{3}+c^{3}-3abc$$ is:

a + b + c = 9

ab + bc + ca = -22

We need to find the value of a³ + b³ + c³- 3abc

We know,

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

Now we need to find the value of a² + b² + c²

We also know,

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

Putting the value of a + b + c = 9 and ab + bc + ca = -22

( 9 )² = a² + b² + c² + 2 ( -22 )

81 = a² + b² + c² -44

a² + b² + c² = 81 + 44

a² + b² + c² = 125

Now,

Substituting value in the formula for a³ + b³ + c³- 3abc

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 ) )

a³ + b³ + c³- 3abc = ( 9 ) ( 125 - ( =22) )

a³ + b³ + c³- 3abc = ( 9 ) ( 147 )

a³ + b³ + c³- 3abc = 1323