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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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:
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