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(D) From equation $I$: $(x+y)^{3} = 1331 \implies x+y = \sqrt[3]{1331} = 11$. So,$y = 11-x$. Substitute this into equation $III$: $x(11-x) = 28 \impli

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if $x = y$ or relationship between $x$ and $y$ cannot be established.

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(D) From equation $I$: $(x+y)^{3} = 1331 \implies x+y = \sqrt[3]{1331} = 11$. So,$y = 11-x$. Substitute this into equation $III$: $x(11-x) = 28 \implies 11x - x^{2} = 28$. Rearranging gives the quadratic equation: $x^{2} - 11x + 28 = 0$. Factoring the quadratic: $x^{2} - 7x - 4x + 28 = 0 \implies x(x-7) - 4(x-7) = 0 \implies (x-7)(x-4) = 0$. Thus,$x = 7$ or $x = 4$. If $x = 7$,then $y = 11-7 = 4$. If $x = 4$,then $y = 11-4 = 7$. Now,using equation $II$: $x-y+z = 0 \implies z = y-x$. Case $1$: If $x = 7, y = 4$,then $z = 4-7 = -3$. Case $2$: If $x = 4, y = 7$,then $z = 7-4 = 3$. Since $x$ can be greater than $y$ or less than $y$,no fixed relationship between $x$ and $y$ can be established.

Solve the given equations and select the correct answer from the given options.
$I. (x+y)^{3} = 1331$
$II. x-y+z = 0$
$III. xy = 28$

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Solve the given equations and select the correct answer from the given options.$I. (x+y)^{3} = 1331$$II. x-y+z = 0$$III. xy = 28$