If `x+y = 7 and xy = 12`, then the vaue of `((1)/(x^(3))+(1)/(y^(3)))` is :

Options

A. `(191)/(1728)`
B. 1
C. `(97)/(1728)`
D. `(91)/(1728)`

Answer

To solve the problem, we need to find the value of \(\frac{1}{x^3} + \frac{1}{y^3}\) given that \(x + y = 7\) and \(xy = 12\).

### Step-by-Step Solution:

1. **Use the identity for the sum of cubes**:
We know that:
\[ x^3 + y^3 = (x+y)(x^2 - xy + y^2) \]
We can also express \(x^2 + y^2\) in terms of \(x+y\) and \(xy\):
\[ x^2 + y^2 = (x+y)^2 - 2xy \]

2. **Calculate \(x^2 + y^2\)**:
Substitute the known values into the equation:
\[ x^2 + y^2 = 7^2 - 2 \cdot 12 = 49 - 24 = 25 \]

3. **Calculate \(x^3 + y^3\)**:
Now substitute \(x+y\) and \(x^2 + y^2\) into the identity:
\[ x^3 + y^3 = (x+y)((x^2 + y^2) - xy) = 7(25 - 12) = 7 \cdot 13 = 91 \]

4. **Find \(x^3y^3\)**:
We can find \(x^3y^3\) using the identity:
\[ x^3y^3 = (xy)^3 = 12^3 = 1728 \]

5. **Calculate \(\frac{1}{x^3} + \frac{1}{y^3}\)**:
We can express this as:
\[ \frac{1}{x^3} + \frac{1}{y^3} = \frac{x^3 + y^3}{x^3y^3} \]
Substitute the values we calculated:
\[ \frac{1}{x^3} + \frac{1}{y^3} = \frac{91}{1728} \]

6. **Final Result**:
Thus, the value of \(\frac{1}{x^3} + \frac{1}{y^3}\) is:
\[ \frac{91}{1728} \]