© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the roots and the given equation
We have the quadratic equation
$ x^2 + (3)^{\tfrac{1}{4}}\,x + \sqrt{3} = 0 $
with distinct roots $ \alpha $ and $ \beta $. We aim to find the value of
$ \alpha^{96}\bigl(\alpha^{12} - 1\bigr) + \beta^{96}\bigl(\beta^{12} - 1\bigr). $
Step 2: Relate $ \alpha^2 $ to $ \alpha $
Since $ \alpha $ is a root, it satisfies
$ \alpha^2 = - (3)^{\tfrac{1}{4}}\,\alpha \;-\; \sqrt{3}. $
We will use this relation to obtain higher powers of $ \alpha $ in terms of simpler expressions.
Step 3: Square appropriately to find a polynomial in $ \alpha $
From
$ \alpha^2 + \sqrt{3} = -(3)^{\tfrac{1}{4}} \,\alpha, $
square both sides:
$ (\alpha^2 + \sqrt{3})^2 \;=\; \bigl((3)^{\tfrac{1}{4}}\,\alpha\bigr)^2. $
Hence
$ \alpha^4 \;+\; 2\,\sqrt{3}\,\alpha^2 \;+\; 3 \;=\; (3)^{\tfrac{1}{2}}\,\alpha^2. $
Rearranging terms eventually gives a polynomial that $ \alpha $ satisfies. With repeated squaring and simplifying, one can show that
$ \alpha^8 + 3\,\alpha^4 + 9 \;=\; 0
\quad \Longrightarrow \quad
\alpha^8 = -\,9 \;-\; 3\,\alpha^4.
Step 4: Express $ \alpha^{12} $ in simpler terms
Multiplying $ \alpha^8 = -9 - 3\,\alpha^4 $ by $ \alpha^4 $ gives
$ \alpha^{12} = -\,9\,\alpha^4 \;-\; 3\,\alpha^8.
Then substitute $ \alpha^8 = -9 - 3\,\alpha^4 $ again:
$ \alpha^{12}
\;=\;
-9\,\alpha^4 \;-\; 3\bigl(-9 \;-\; 3\,\alpha^4\bigr)
\;=\;
-9\,\alpha^4 + 27 + 9\,\alpha^4
\;=\;
27.
Thus we obtain
$ \alpha^{12} = 27. $
(Notice $ 27 = 3^3. $)
Step 5: Find $ \alpha^{96} $
We have
$ \alpha^{12} = 27. $
Hence
$ (\alpha^{12})^8 = 27^8
\;\Longrightarrow\;
\alpha^{96} = 27^8.
Since $ 27 = 3^3, $
we get
$ 27^8 = (3^3)^8 = 3^{24}, $
so
$ \alpha^{96} = 3^{24}.
Step 6: Compute the expression for both roots
By symmetry, $ \beta $ (the other root) satisfies the same kind of relations, so
$ \beta^{12} = 27 $
and
$ \beta^{96} = 3^{24}.
$
Step 7: Substitute into the target expression
We want
$ \alpha^{96}(\alpha^{12} - 1) + \beta^{96}(\beta^{12} - 1). $
Substituting the values found:
$ \alpha^{12} = 27 $ and $ \alpha^{96} = 3^{24}.$
$ \beta^{12} = 27 $ and $ \beta^{96} = 3^{24}.$
Hence
$ \alpha^{96}(\alpha^{12} - 1) + \beta^{96}(\beta^{12} - 1)
\;=\;
3^{24}(27 - 1) + 3^{24}(27 - 1)
\;=\;
3^{24} \times 26 \;+\; 3^{24} \times 26
\;=\;
2 \times 26 \,\times\, 3^{24}
\;=\;
52 \times 3^{24}.
Step 8: Conclude the correct answer
Therefore, the required value is
$ 52 \times 3^{24}. $
This matches Option (3).
