Question : If $x^{2}+\frac{1}{x^{2}}=1$, then, what is the value of $x^{48}+x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^{6}+1$?
Option 1: – 9
Option 2: 0
Option 3: 1
Option 4: 9
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Correct Answer: 1
Solution :
Given: $x^{2}+\frac{1}{x^{2}} = 1$
Adding 2 on both sides, we get,
$x^{2} + \frac{1}{x^{2}} + 2 = 1 + 2$
⇒ $x^{2} + \frac{1}{x^{2}} + 2×x^2×\frac{1}{x^{2}} = 3$
⇒ $(x + \frac{1}{x})^{2} = 3$
$x + \frac{1}{x} = \sqrt3$
Cubing both sides, we get,
⇒ $(x + \frac{1}{x})^{3} = (\sqrt3)^2$
⇒ $x^{3} + \frac{1}{x^{3}} + 3x\times\frac{1}{x}(x + \frac{1}{x})= 3\sqrt3$
⇒ $x^{3} + \frac{1}{x^{3}} = 0$
⇒ $x^{3} = – \frac{1}{x^{3}}$
⇒ $x^{6} = -1$
Putting values of x
6
= –1,
$x^{48} + x^{42} + x^{36} + x^{30} + x^{24} + x^{18} + x^{12} + x^{6} + 1$
= $1 - 1 + 1 -1 + 1 -1 + 1 -1 + 1$
= 1
Hence, the correct answer is 1.
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