The value of:
`(9^(x)(9^(x-1)))^(x)/(9^(x+1).3^(2x-2)){(729^(x/3))/81}^(-x) + (3^(a) -2^(3) .3^(a-2))/(3^(a) -3^(a-1))` is:

Options

A. 9
B. 6
C. 12
D. 16

Answer

To solve the expression

\[ \frac{(9^{x}(9^{x-1}))^{x}}{9^{x+1} \cdot 3^{2x-2}} \cdot \left(\frac{(729^{\frac{x}{3}})}{81}\right)^{-x} + \frac{(3^{a} - 2^{3} \cdot 3^{a-2})}{(3^{a} - 3^{a-1})} \]

we will break it down step by step.

### Step 1: Simplify the numerator of the first term

The numerator is \((9^{x}(9^{x-1}))^{x}\).

Using the property of exponents \(a^m \cdot a^n = a^{m+n}\):

\[ 9^{x} \cdot 9^{x-1} = 9^{x + (x-1)} = 9^{2x - 1} \]

Now, raise this to the power of \(x\):

\[ (9^{2x - 1})^{x} = 9^{(2x - 1)x} = 9^{2x^2 - x} \]

### Step 2: Simplify the denominator of the first term

The denominator is \(9^{x+1} \cdot 3^{2x-2}\).

Since \(9 = 3^2\), we can rewrite \(9^{x+1}\):

\[ 9^{x+1} = (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x + 2} \]

Thus, the denominator becomes:

\[ 3^{2x + 2} \cdot 3^{2x - 2} = 3^{(2x + 2) + (2x - 2)} = 3^{4x} \]

### Step 3: Combine the first term

Now we can rewrite the first term:

\[ \frac{9^{2x^2 - x}}{3^{4x}} \]

Since \(9 = 3^2\), we can express \(9^{2x^2 - x}\) as:

\[ (3^2)^{2x^2 - x} = 3^{2(2x^2 - x)} = 3^{4x^2 - 2x} \]

Now, substituting this back into our expression:

\[ \frac{3^{4x^2 - 2x}}{3^{4x}} = 3^{(4x^2 - 2x) - 4x} = 3^{4x^2 - 6x} \]

### Step 4: Simplify the second term

The second term is:

\[ \frac{(3^{a} - 2^{3} \cdot 3^{a-2})}{(3^{a} - 3^{a-1})} \]

Factor out \(3^{a-2}\) from the numerator:

\[ 3^{a-2}(3^2 - 2^3) = 3^{a-2}(9 - 8) = 3^{a-2}(1) = 3^{a-2} \]

Now, factor out \(3^{a-1}\) from the denominator:

\[ 3^{a-1}(3 - 1) = 3^{a-1}(2) \]

Thus, the second term simplifies to:

\[ \frac{3^{a-2}}{2 \cdot 3^{a-1}} = \frac{3^{a-2}}{2 \cdot 3^{a-1}} = \frac{1}{2} \cdot \frac{3^{a-2}}{3^{a-1}} = \frac{1}{2} \cdot 3^{-1} = \frac{1}{6} \]

### Step 5: Combine both terms

Now we have:

\[ 3^{4x^2 - 6x} + \frac{1}{6} \]

### Final Answer

The value of the expression is:

\[ 3^{4x^2 - 6x} + \frac{1}{6} \]

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