If $a + b + c = 19, ab + bc + ca = 120$, then what is the value of $a^3 + b^3 + c^3 - 3abc$?
Answer & explanation
Correct answer: option 4
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