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If \( \Large a^{2} + 1 = a \), then what is the value of \( \Large a^{12} + a^{6} + 1 \) is
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A) 2 |
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B) 3 |
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C) -3 |
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D) 1 |
Correct Answer:
| B) 3 |
Description for Correct answer:
\( \Large a^{2} + 1 => a = a + \frac{1}{a} = 1 \)
On squaring both sides, we get
\( \Large a^{2} + \frac{1}{a^{2}}= -1 \)
On cubing both sids, we gt
\( \Large \left(a^{2} + \frac{1}{a^{2}}\right)^{3} = \left(-1\right)^{3} \)
=> \( \Large a^{6} + \frac{1}{a^{6}} + 3a^{2} \times \frac{1}{a^{2}} \left(a^{2} + \frac{1}{a^{2}}\right) = -1 \)
=> \( \Large a^{6} + \frac{1}{a^{6}} + 3 \times \left(-1\right) = -1 \)
Now, \( \Large a^{6} + \frac{1}{a^{6}} + 1 = 3 \)
As \( \Large a^{12} + a^{6} + 1 \) can also be written as
\( \Large a^{6} + \frac{1}{a^{6}} + 1 \)
Therefore, \( \Large a^{12} + a^{6} + 1 = 3 \)
Part of solved Simplification questions and answers : >> Elementary Mathematics >> Simplification
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