If x2-x-1-0 then the value of x-1-x2-x2-1-x22-x3-1-x32-x5-1-x52 is

Given : x^2 x+1=0 ⇒ x+1x =1 ⇒ x^2+1x^2 +2=1 ⇒ nbsp; nbsp;x^2+1x^2 = 1 Cubing x+1x =1, we get x^3+1x^3+3(x+ 1 x) = 1 ⇒ nbsp;x^3+1x^3 = 2 Also, (x^2+1x^2) ...

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Byju's Answer

Standard XII

Mathematics

Quadratic Polynomial

If x2-x+1=0 t...

Question

If $x^{2} - x + 1 = 0$ then the value of ${(x + \frac{1}{x})}^{2} + {(x^{2} + \frac{1}{x^{2}})}^{2} + {(x^{3} + \frac{1}{x^{3}})}^{2} + {(x^{5} + \frac{1}{x^{5}})}^{2}$ is

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Solution

Given : $x^{2} - x + 1 = 0$
$\Rightarrow x + \frac{1}{x} = 1 \Rightarrow x^{2} + \frac{1}{x^{2}} + 2 = 1 \Rightarrow x^{2} + \frac{1}{x^{2}} = - 1$
Cubing $x + \frac{1}{x} = 1$ , we get
$x^{3} + \frac{1}{x^{3}} + 3 (x + \frac{1}{x}) = 1 \Rightarrow x^{3} + \frac{1}{x^{3}} = - 2$

Also,
$(x^{2} + \frac{1}{x^{2}}) \cdot (x^{3} + \frac{1}{x^{3}}) = x^{5} + \frac{1}{x^{5}} + x + \frac{1}{x} \Rightarrow 2 = x^{5} + \frac{1}{x^{5}} + 1 \Rightarrow x^{5} + \frac{1}{x^{5}} = 1$

Therefore,
${(x + \frac{1}{x})}^{2} + {(x^{2} + \frac{1}{x^{2}})}^{2} + {(x^{3} + \frac{1}{x^{3}})}^{2} + {(x^{5} + \frac{1}{x^{5}})}^{2} = 1^{2} + (- 1)^{2} + (- 2)^{2} + 1^{2} = 7$

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If x2−x+1=0 then the value of (x+1x)2+(x2+1x2)2+(x3+1x3)2+(x5+1x5)2 is

Given : x2−x+1=0 ⇒x+1x=1⇒x2+1x2+2=1⇒x2+1x2=−1 Cubing x+1x=1, we get x3+1x3+3(x+1x)=1⇒x3+1x3=−2 Also, (x2+1x2)⋅(x3+1x3)=x5+1x5+x+1x⇒2=x5+1x5+1⇒x5+1x5=1 Therefore, (x+1x)2+(x2+1x2)2+(x3+1x3)2+(x5+1x5)2=12+(−1)2+(−2)2+12=7

If $x^{2} - x + 1 = 0$ then the value of ${(x + \frac{1}{x})}^{2} + {(x^{2} + \frac{1}{x^{2}})}^{2} + {(x^{3} + \frac{1}{x^{3}})}^{2} + {(x^{5} + \frac{1}{x^{5}})}^{2}$ is