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If \$\$x^4+x^3+x^2+x+1=0\$\$ then what"s the value of \$x^5\$ ??

I thought it would be \$-1\$ but it does not satisfy the equation

\$x^4+x^3+x^2+x+1 = 0\$

Multiply everything by \$x\$

\$x^5+x^4+x^3+x^2+x = 0\$

Add \$1\$ to both sides:

\$x^5+x^4+x^3+x^2+x+1 = 1\$

\$x^5+(x^4+x^3+x^2+x+1) = 1\$

\$x^5+(0) = 1\$

\$x^5=1\$

The left side of the equation is \$P= (x^4 +x^2) + (x^3 + x) +1 = 2\$

In the equation \$x^2(x^2+1) +1 > 0\$, there is no such real number for \$x\$. Logically, a false statement implies anything, so \$x^5\$ = anything you wish .

M Andrews" correct argument above can be modified to lớn see that \$x^5-1 =(x-1)(x^4+x^3+x^2+x+1)\$, so the roots of \$P=0\$ are the 4 other complex 5th roots of unity (which are not x=1) namely \$x=e^2mpi i/5\$, so \$m=1,2,3,4\$, và so on.

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For instance, for \$m=1\$ \$x= cos(2pi/5) + i sin(2pi/5)\$ . So \$x^5 =1 \$, provided you allow complex numbers as roots of P.

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