Method (1) Given ,P(x)=f(x3)+xg(x3)⋅⋅⋅⋅⋅⋅⋅(i) ∴P(1)=f(1)+g(1)⋅⋅⋅⋅⋅⋅⋅(ii) Given, P(x) is divisible by (x2+x+1). ∴P(x)=Q(x).(x2+x+1) As, we know that ω and ω2 are non-real cube roots of unity and this is also root of x2+x+1=0 ∴P(ω)=P(ω2)=0 As, we know that ω and ω2 are non-real cube roots of unity and this is also root of x2+x+1=0 ∴P(ω)=P(ω2)=0... (iii) From Eq. (i), P(ω)=f(ω3)+ω[g(ω)3]=0[ from Eq. (iii) ] ⇒f(1)+ωg(1)=0... (iv) and P(ω2)=0 [from Eq. (iii)] ⇒f(ω6)+ω2⋅g(ω6)=0 ⇒f(1)+ω2g(1)=0⋅⋅⋅⋅⋅⋅⋅(v) Now, adding Eqs. (iv) and (v), we get 2f(1)+(ω+ω2)g(1)=0 ⇒2f(1)−1g(1)=0(∵1+ω+ω2=0) ⇒2f(1)=g(1)... (vi) Subtracting Eq. (iv) from Eq. (v), we get 0+(ω−ω2)g(1)=0 ⇒g(1)=0 f(1)=
g(1)
2
=
0
2
[ from Eq. (vi) ] ⇒f(1)=0 From Eq. (ii), P(1)=f(1)+g(1)=0+0=0 Method (2) ∵P(ω)=0 ⇒f(1)+ωg(1)=0 ⇒f(1)+(−
1
2
+
√3i
2
)g(1)=0 ⇒(f(1)−
g(1)
2
)+i(
√3
2
g(1))=0 On comparing real and imaginary parts from both sides, we have llf(1)−