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Question 72
If x + y + z = 19, xyz = 216 and xy + yz + zx = 114, then the value of $$\sqrt{x^3 + y^3 + z^3 + xyz}$$ is:
Solution
$$we\ have,\ x+y+z=19$$
$$xy+yz+zx=114$$
$$xyz=216$$
$$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
$$361=x^2+y^2+z^2+228$$
$$we\ get\ ,\ x^2+y^2+z^2=133$$
$$Now,\ x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
$$19\left(133-114\right)+3\times\ 216$$
$$361+648\ =\ 1009$$
$$we\ get\ ,\ x^3+y^3+z^3=1009$$
$$Now\sqrt{\ x^3+y^3+z^3+xyz}=\sqrt{\ 1009+216}\ =\ 35$$
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