JEE Main 2021 — Mathematics Algebra

$$P(x)=f({x}^{3})+xg({x}^{3})$$P(x)=f(x3)+xg(x3)

$$P(1)=f(1)+g(1)...(1)$$P(1)=f(1)+g(1)...(1)

Now $P(x)$P(x) is divisible by ${x}^{2}+x+1$x2+x+1

$$\Rightarrow P(x)=Q(x)({x}^{2}+x+1)$$⇒P(x)=Q(x)(x2+x+1)

$P(\omega )=0=P({\omega }^{2})$P(ω)=0=P(ω2) where $\omega ,{\omega }^{2}$ω,ω2 are non-real cube roots of units

$$P(x)=f({x}^{3})+xg({x}^{3})$$P(x)=f(x3)+xg(x3)

$$P(\omega )=f({\omega }^{3})+\omega g({\omega }^{3})=0$$P(ω)=f(ω3)+ωg(ω3)=0

$$f(1)+\omega g(1)=0...(2)$$f(1)+ωg(1)=0...(2)

$$P({\omega }^{2})=f({\omega }^{6})+{\omega }^{2}g({\omega }^{6})=0$$P(ω2)=f(ω6)+ω2g(ω6)=0

$$f(1)+{\omega }^{2}g(1)=0...(3)$$f(1)+ω2g(1)=0...(3)

Now, $(2)+(3)$(2)+(3)

$$\Rightarrow 2f(1)+(\omega +{\omega }^{2})g(1)=0$$⇒2f(1)+(ω+ω2)g(1)=0

$$2f(1)=g(1)...(4)$$2f(1)=g(1)...(4)

Now, $(2)-(3)$(2)−(3)

$$\Rightarrow (\omega -{\omega }^{2})g(1)=0$$⇒(ω−ω2)g(1)=0

$g(1)=0=f(1)$g(1)=0=f(1) from $(4)$(4)

from $(1),$(1), $P(1)=f(1)+g(1)=0$P(1)=f(1)+g(1)=0