Question : If $x+y+z=19, x y z=216$ and $x y+y z+z x=114$, then the value of $x^3+y^3+z^3+x y z$ is:
Option 1: 1225
Option 2: 1441
Option 3: 361
Option 4: 577
Correct Answer: 1225
Solution :
$x + y + z = 19$ --------------(1)
$xyz = 216$ --------------------(2)
$xy + yz + zx= 114$ -----------(3)
Squaring both sides in equation (1)
$(x + y + z)^2 = 19^2$
$⇒x^2 + y^2 + z^2 + 2(xy + yz + zx) = 361$
$⇒x^2 + y^2 + z^2 = 361 - 2 × 114$
$⇒x^2 + y^2 + z^2 = 133$
Now, $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$
$⇒x^3 + y^3 + z^3= 19(133 - 114) + 648$
$⇒x^3 + y^3 + z^3 = 19 × 19 + 648$
$⇒x^3 + y^3 + z^3= 1009$
So, $x^3 + y^3 + z^3 + xyz= 1009 + 216= 1225$
Hence, the correct answer is 1225.
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