Let x, y, z be non-zero real numbers such that `x/y + y/z + z/x = 7 `and `y/x + z/y + x/z = 9 , ` then `(x^3)/y^3 + y^3/z^3 + z^3/x^3 - 3` is equal to

Options

A. 152
B. 153
C. 154
D. 155

Answer

To solve the problem, we need to find the value of the expression \((x^3/y^3 + y^3/z^3 + z^3/x^3 - 3)\) given the equations:

1. \(\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 7\)
2. \(\frac{y}{x} + \frac{z}{y} + \frac{x}{z} = 9\)

Let's denote:
- \(a = \frac{x}{y}\)
- \(b = \frac{y}{z}\)
- \(c = \frac{z}{x}\)

From the definitions, we have:
- \(abc = \frac{x}{y} \cdot \frac{y}{z} \cdot \frac{z}{x} = 1\)

Now we can rewrite the given equations in terms of \(a\), \(b\), and \(c\):
1. \(a + b + c = 7\)
2. \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 9\)

We can rewrite the second equation as:
\[ \frac{bc + ca + ab}{abc} = 9 \implies bc + ca + ab = 9 \cdot 1 = 9 \]

Now we have the following system of equations:
1. \(a + b + c = 7\)
2. \(ab + bc + ca = 9\)

We need to find \(a^3 + b^3 + c^3 - 3\).

Using the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)((a + b + c)^2 - 3(ab + ac + bc)) \]

Substituting the known values:
- \(a + b + c = 7\)
- \(ab + ac + bc = 9\)
- \(abc = 1\)

Now we compute:
\[ a^3 + b^3 + c^3 - 3 = (a + b + c)((a + b + c)^2 - 3(ab + ac + bc)) + 3abc - 3 \]

Calculating \((a + b + c)^2\):
\[ (7)^2 = 49 \]

Now substituting into the identity:
\[ a^3 + b^3 + c^3 - 3 = 7(49 - 3 \cdot 9) + 3 \cdot 1 - 3 \]
\[ = 7(49 - 27) + 3 - 3 \]
\[ = 7(22) + 0 \]
\[ = 154 \]

Thus, the final answer is:
\[ \boxed{154} \]