If ` sin^(-1) x + sin^(-1) y + sin^(-1) z = (3pi)/2 ",then find the value of " Sigma ((x^(101) + y ^(101))(x^(202)+y^(202)))/((x^(303)+y^(303))(x^(404)+y^(404)))`

Answer

To solve the problem, we start with the equation given:

\[ \sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2} \]

### Step 1: Analyze the equation
The equation \(\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2}\) implies that each of the inverse sine functions must equal \(\frac{\pi}{2}\) because the maximum value of \(\sin^{-1} t\) for \(t \in [-1, 1]\) is \(\frac{\pi}{2}\). Thus, we can conclude:

\[ \sin^{-1} x = \sin^{-1} y = \sin^{-1} z = \frac{\pi}{2} \]

### Step 2: Find the values of \(x\), \(y\), and \(z\)
From the above conclusion, we can find the values of \(x\), \(y\), and \(z\):

\[ x = \sin\left(\frac{\pi}{2}\right) = 1 \]
\[ y = \sin\left(\frac{\pi}{2}\right) = 1 \]
\[ z = \sin\left(\frac{\pi}{2}\right) = 1 \]

### Step 3: Substitute the values into the expression
Now we need to evaluate the expression:

\[ \Sigma \frac{(x^{101} + y^{101})(x^{202} + y^{202})}{(x^{303} + y^{303})(x^{404} + y^{404})} \]

Substituting \(x = 1\) and \(y = 1\):

\[ \Sigma \frac{(1^{101} + 1^{101})(1^{202} + 1^{202})}{(1^{303} + 1^{303})(1^{404} + 1^{404})} \]

### Step 4: Simplify the expression
Now we simplify each term:

\[ 1^{101} + 1^{101} = 1 + 1 = 2 \]
\[ 1^{202} + 1^{202} = 1 + 1 = 2 \]
\[ 1^{303} + 1^{303} = 1 + 1 = 2 \]
\[ 1^{404} + 1^{404} = 1 + 1 = 2 \]

Substituting these values back into the expression:

\[ \Sigma \frac{(2)(2)}{(2)(2)} = \Sigma \frac{4}{4} = \Sigma 1 \]

### Step 5: Count the number of terms in the summation
Since we have three variables \(x\), \(y\), and \(z\), the summation will have three terms:

\[ 1 + 1 + 1 = 3 \]

### Final Answer
Thus, the value of the expression is:

\[ \boxed{3} \]