If x2+ x +1=0 then find the value of (x3+1by x)3

2
if p(x) =x4+ax3+bx2+cx+d and p(1)=p(2)=p(3)=0 then find the value p(4)+p(0)
simple method x^2+x+1=0 so taking x common we get x+1/x=-1 now
what?
x^3+1/x^3=(x+1/x)(x^2-1+1/x^2+ (+2-2)=(x+1/x)*[(x+1/x)^2-3]=(-1)(1-3)=2
As no place for new answer, let me add it as comment....
as x^2+x+1=0 it is obvious that x is not equal to 1.
x^2+x+1=0
(x-1)(x^2+x+1)=0
x^3-1=0
x^3=1
so x is the cube root of unity,
x = 1, w, w^2... here w is omega
now x^3+1 = w^3 + 1 = 1 + 1 = 2
and 1/w = w^2
So, ( (x^3+1)/x )^3 = (2 w^2)^3 = 8 w^6 = 8 (w^3)^2 = 8

this is the correct way of solving...

Answers

2014-10-30T12:46:41+05:30

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So given 

x² + x + 1 = 0 ⇒ x² + x + 1 - 1 + 1 = 0 ⇒ x² + x - 1 = -2

so
{\frac{x^3+1}{x}}^3


so x(x² + x + 1) = x³ + x² + x + 1 - 1

x² + x + 1 = (x³ + 1)/x  + (x² + x - 1)/x

0 = (x³ + 1)/x  + (x² + x - 1)/x

- (x² + x - 1)/x = (x³ + 1)/x  

- ( - 2) = (x³ + 1)

x³ + 1 = 2 



so value for {(\frac{x^3+1}{x})}^3=(2^3/x^3)=8
1 5 1
hope it helps
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You've got x=1, but does it satisfy the given equation
of course x=1, is not the solution
2014-10-30T13:55:50+05:30

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x\ is\ like\ cube\ root\ of\ 1,\ \omega.\\\\(x-1)(1+x+x^2)=0\\\\x^3-1=0,\ \ ==> \ x^3=1\\\\(\frac{x^3+1}{x})^3=\frac{(x^3+1)^3}{x^3}=\frac{(1+1)^3}{1}=8\\\\
3 3 3
you are right, x is cube root of unity, but x != 1, so you should solve it by taking x = omega, not 1....
x != 1 , that is why we multiply 1+x+x^2 with (x-1)... 1+x+x^2 = 0 has roots omega and omega^2. they are the complex cube roots of 1.