1as it is divisible by x2+x+1 we can write
xf(x3)+x2g(x6) = (x2+x+1)H(x)
where H(x) is another polynomial
as this is valid for all x
put x=ω
ωf(1)+ω2g(1)=0
put x=ω2
ω2f(1)+ω4g(1)=0
solving we get f(1)=g(1)=0
If f(x) and g(x) are two polynomial such that the polynomial h(x) = xf(x3) + x2g(x6) is divisible by x2+x+1, then
a. f(1) = g(1)
b. f(1) = g(1) ≠0
c. f(1) = -g(1) ≠0
d. none of these
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