If $a + b + c = 19, ab + bc + ca = 120$, then what is the value of $a^3 + b^3 + c^3 - 3abc$?

19

If $a + b + c = 19

ab + bc + ca = 120$

Then what is the value of $a^3 + b^3 + c^3 - 3abc$?

If the number of equations are less than the number of variables then we can put the extra variables according to our choice = 

So here two equations given and three variables are present so put c = 0

If $a + b = 19

ab  = 120$

Then what is the value of $a^3 + b^3 $?

If x + y  = n

then, $x^3 + y^3$ = n³ - 3 × n × xy

$a^3 + b^3 $ = 19³ - 3 × 19 × 120

$a^3 + b^3 $ = 6859 - 6840 = 19