If $a^3 + b^3 + c^3 - 3abc = 250 $ and a + b + c = 10, then what will be the value of $\frac{1}{5} (ab + bc + ca)$ ?
Answer & explanation
Correct answer: option 2
If $a^3 + b^3 + c^3 - 3abc = 250 $
a + b + c = 10,
then what will be the value of $\frac{1}{5} (ab + bc + ca)$ = ?
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^3 + b^3= 250 $
a + b = 10,
then what will be the value of $\frac{1}{5} (ab)$ = ?
(a + b)3 = a3 + b3 + 3ab(a+b)
(10)3 = 250 + 3ab(10)
1000 = 250 + 30ab
30ab = 750
ab = 25
and $\frac{1}{5} (ab)$ = $\frac{1}{5} (25)$ = 5