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

Question

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

Team Careers360 · Accepted Answer

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,
⇒ $(x + \frac{1}{x})^{3} = (\sqrt3)^2$
⇒ $x^{3} + \frac{1}{x^{3}} + 3x imes\frac{1}{x}(x + \frac{1}{x})= 3\sqrt3$
⇒ $x^{3} + \frac{1}{x^{3}} = 0$
⇒ $x^{3} = – \frac{1}{x^{3}}$
⇒ $x^{6} = -1$
Putting values of x⁶= –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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Question : If x2+1x2=1, then, what is the value of x48+x42+x36+x30+x24+x18+x12+x6+1?Option 1: – 9Option 2: 0Option 3: 1Option 4: 9

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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