Question : If $x+y+z=19, x y z=216$ and $x y+y z+z x=114$, then the value of $\sqrt{x^3+y^3+z^3+x y z}$ is:
Option 1: 32
Option 2: 30
Option 3: 28
Option 4: 35
Correct Answer: 35
Solution :
We know,
$x^3 + y^3 + z^3 = (x + y + z)[(x + y + z)^2 - 3(xy + yz + zx) + 3xyz]$
Now, We have-
$\sqrt{x^3 + y^3 + z^3 + xyz}$
$=\sqrt{(x + y + z)[(x + y + z)^2 - 3(xy + yz + zx) + 3xyz+ xyz]}$
$=\sqrt{19 × (19^2 - 3 × 114) + 4 × 216}$
$=\sqrt{19(361 - 342) + 864}$
$=\sqrt{19 × 19 + 864}$
$=\sqrt{1225}$
$=35$
Hence, the correct answer is 35.
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