Question 73

if x+ $$\frac{1}{x}$$=$$\surd{3}$$ then the value of $$x^{18}+x^{12}+x^{6}+1$$

Solution

Given that x+ $$\frac{1}{x}$$=$$\surd{3}$$

Squaring on both sides, we get

$$(x+ \frac{1}{x})^{3}=(\surd{3})^{3}$$

=> $$x^{3}+\frac{1}{x^3}+3\surd{3}=3\surd{3}$$

=> $$x^{3}+\frac{1}{x^3} = 0 $$

=> $$x^{3}= - \frac{1}{x^3} $$

=> $$x^{6}= -1 $$

Squaring on both sides

=> $$x^{12}= 1 $$

$$ (x^{6})^{3} = (-1)^{3} = -1 $$

Therefore,

$$x^{18}+x^{12}+x^{6}+1$$ = $$ -1 + 1 -1 + 1 = 0 $$

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