Question

X+y+z=19 xyz=216 xy+yz+zx=114 √x^3+y^3+z^3+xyz

Answer 1

Answer:

$\sqrt{{x}^{3} + {y}^{3} + {z}^{3} + xyz} = 25.59$

Explanation:

As we know that

${x}^{3} + {y}^{3} + {z}^{3} - 3xyz = (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - xz)$

To find the value of

$\sqrt{ {x}^{3} + {y}^{3} + {z}^{3} + xyz }$

Add 4xyz in both side of the formula

${x}^{3} + {y}^{3} + {z}^{3} - 3xyz + 4xyz =4xyz + (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - xz) \\ \\ {x}^{3} + {y}^{3} + {z}^{3} + xyz =4xyz + (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - xz)$

We don't have the value of

${x}^{2} + {y}^{2} + {z}^{2}$

for that square both sides

$x + y + z = 19 \\ \\ {(x + y + z)}^{2} = 361 \\ \\ {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx = 361 \\ \\ {x}^{2} + {y}^{2} + {z}^{2} = 361 - 2(xy + yz + zx) \\ \\ {x}^{2} + {y}^{2} + {z}^{2} = 361 - 2(114) \\ \\ {x}^{2} + {y}^{2} + {z}^{2} = 133$

Now place all the values

${x}^{3} + {y}^{3} + {z}^{3} + xyz =4(216) + (19)( 133- 144) \\ \\ {x}^{3} + {y}^{3} + {z}^{3} + xyz= 864 - 209 \\ \\ {x}^{3} + {y}^{3} + {z}^{3} + xyz = 655$

Now,take square root both sides

$\sqrt{{x}^{3} + {y}^{3} + {z}^{3} + xyz} = \sqrt{655} \\ \\ \sqrt{{x}^{3} + {y}^{3} + {z}^{3} + xyz} = 25.59$

Hope it helps you.