If $l o g_{x} y, l o g_{z} x, l o g_{y} z$ are in G.P., $x y z = 64$ and $x^{3}, y^{3}, z^{3}$ are in A.P., find $x + y + z$

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Solution

According to question,
$\frac{l o g_{z} x}{l o g_{x} y} = \frac{l o g_{y} z}{l o g_{z} x} \Rightarrow (l o g x)^{3} = (l o g z)^{3}$
$\Rightarrow x = z$
Since $2 y^{3} = x^{3} + z^{3} \Rightarrow x^{2} = y^{3}$ or $x = y$
Given $x y z = 64$ & $x = y = z$
$∴ x = y = z = 4$
& $x + y + z = 12$

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Similar questions

x^{3}, y^{3}, z^{3}

A . P .

{l o g}_{x} y, {l o g}_{z} x, {l o g}_{y} z

G . P .

x y z = 64

x = y = z = 4

\frac{1}{l o g_{x y} x y z} + \frac{1}{l o g_{y z} x y z} + \frac{1}{l o g_{z x} x y z} =

x + y + z = 0

\frac{x^{3} + y^{3}}{x y z} = \frac{y^{3} + z^{3}}{x y z} = \frac{x^{3} + z^{3}}{x y z} = a

x = \sqrt{2} - 1, y = 1 - \sqrt{3}

z = \sqrt{3} - \sqrt{2}

x^{3} + y^{3} + z^{3} - x y z .

x + y + z = 1, x y + y z + z x = - 1

x y z = - 1

x^{3} + y^{3} + z^{3}

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