Question

if x + 1/x = 1 find, x¹⁵ + x¹² + x⁹ + x⁶ + x³ + 1​

Answer 1

Given that,

$\rm{x+\dfrac{1}{x}=1}$

$\rm{x^{2}-x+1=0}$

which shares the roots to the cubic equation

$\rm{(x+1)(x^{2}-x+1)=0}$

$\rm{x^{3}+1=0}$

$\rm{\therefore x^{3}=-1}$

Hence,

$\small\rm{ x^{15}+x^{12}+x^{9}+x^{6}+x^{3}+1=(-1)^{5}+(-1)^{4}+(-1)^{3}+(-1)^{2}+(-1)^{1}+1 }$

$\rm{\therefore x^{15}+x^{12}+x^{9}+x^{6}+x^{3}+1=0}$

Answer 2

Step-by-step explanation:

Given that,

\rm{x+\dfrac{1}{x}=1}x+

x

1

=1

\rm{x^{2}-x+1=0}x

2

−x+1=0

which shares the roots to the cubic equation

\rm{(x+1)(x^{2}-x+1)=0}(x+1)(x

2

−x+1)=0

\rm{x^{3}+1=0}x

3

+1=0

\rm{\therefore x^{3}=-1}∴x

3

=−1

Hence,

\small\rm{$x^{15}+x^{12}+x^{9}+x^{6}+x^{3}+1=(-1)^{5}+(-1)^{4}+(-1)^{3}+(-1)^{2}+(-1)^{1}+1$}

\rm{\therefore x^{15}+x^{12}+x^{9}+x^{6}+x^{3}+1=0}∴x

15

+x

12

+x

9

+x

6

+x

3

+1=0