Question

If x2+(1/x2)=1 find x48+x42+x36+x30+x24+x18+x12+x6+1

Answer 1

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Answer 2

Answer:

Value of $x^{48}+x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1\:is\:1$

Step-by-step explanation:

Given: $x^2+\frac{1}{x^2}=1$

To find: $x^{48}+x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1$ ..........(A)

Consider,

$x^2+\frac{1}{x^2}=1$

$x^4+1=x^2$

$x^4-x^2+1=0$ ................(1)

Now, Multiply (1) with $x^{44}$ we get,

$x^{48}-x^{46}+x^{44}=0\:\:\implies\:\:x^{48}=x^{46}-x^{44}$ .........(2)

Now, Multiply (1) with $x^{32}$ we get,

$x^{36}-x^{34}+x^{32}=0\:\:\implies\:\:x^{36}=x^{34}-x^{32}$ .........(3)

Now, Multiply (1) with $x^{20}$ we get,

$x^{24}-x^{22}+x^{20}=0\:\:\implies\:\:x^{24}=x^{22}-x^{20}$ .........(4)

Now, Multiply (1) with $x^{8}$ we get,

$x^{12}-x^{10}+x^{8}=0\:\:\implies\:\:x^{12}=x^{10}-x^{8}$ .........(5)

Put Value from (2) , (3) , (4) , (5) in (A) and using (1) we get

$x^{48}+x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1$

$=(x^{46}-x^{44}+x^{42})+(x^{34}-x^{32}+x^{30})+(x^{22}-x^{20}+x^{18})+(x^{10}-x^{8}+x^6)+1$

$=x^{42}(x^{4}-x^{2}+1)+x^{30}(x^{4}-x^{2}+1)+x^{18}(x^{4}-x^{2}+1)+x^6(x^{4}-x^{1}+1)+1$

$=x^{42}0)+x^{30}(0)+x^{18}(0)+x^6(0)+1$

= 1

Therefore, Value of $x^{48}+x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1\:is\:1$