Question

X+y+z=19 ,xyz=216 ,xy+yz+zx=114 find $\sqrt{ {x}^{3} + {y}^{3 + {z}^{3} }$​

Answer 1

The answer would be √1009

Step-by-step explanation:

We know that,

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$    .....(1)

∵  $(x+y+z)^2=(x+y)^2 + z^2 + 2(x+y)z$

$=x^2+y^2+z^2+2xy+2yz+2zx$

$\implies x^2+y^2+z^2 = (x+y+z)^2 - 2xy-2yz-2zx$    ......(2),

From equation (1) and (2),$x^3+y^3+z^3-3xyz=(x+y+z)((x+y+z)^2 - 2xy-2yz-2zx-xy-yz-zx)$ $x^3+y^3+z^3-3xyz=(x+y+z)((x+y+z)^2 - 3xy-3yz-3zx)$  $x^3+y^3+z^3=(x+y+z)((x+y+z)^2 - 3(xy+yz+zx))+3xyz$  

We have,

x+y+z=19, xy+yz+zx=114, xyz= 216$x^3+y^3+z^3=(19)((19)^2 - 3(114))+3(216)=19(361-342)+648=19(19)+648=361+648=1009$ Hence,

$\sqrt{x^3+y^3+z^3}=\sqrt{1009}$

#Learn more:

Solve the expression :

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