Omega বিষয়ক

x=p+q,y=p+ x=\mathrm{p}+\mathrm{q}, y=\mathrm{p}+ ωq,z=p+ω2q \omega q, z=p+\omega^{2} q হলে x3+y3+z3x^{3}+y^{3}+z^{3}=?

Solve: দেওয়া আছে, x=p+q,y=p+ x=\mathrm{p}+\mathrm{q}, y=\mathrm{p}+ ωq,z=p+ω2q \omega q, z=p+\omega^{2} q

x3+y3+z3=(p+q)3+(p+ωq)3+(p+ω2q)3=p3+q3+3p2q+3pq2+p3+ω3q3+3p2qω)+3pω2q2+p3+ω6q3+3p2qω2+3pq2ω4)=p3+q3+3p2q+3pq2+p3+q3+3p2qω+3pω2q2+p3+q3+3p2qω2+3pq2ω)=3(p3+q3)+3p2q(1+ω+ω2)+3pq2(1+ω+ω2)x3+y3+z3=3(p3+q3) \begin{aligned} & x^{3}+y^{3}+z^{3}=(p+q)^{3}+(p+\omega q)^{3} + & \left(p+\omega^{2} q\right)^{3} \\ = & \left.p^{3}+q^{3}+3 p^{2} q+3 p q^{2}+p^{3}+\omega^{3} q^{3}+3p^{2} q \omega\right) \\ + & \left.3 p \omega^{2} q^{2}+p^{3}+\omega^{6} q^{3}+3 p^{2} q \omega^{2}+3p q^{2} \omega^{4}\right) \\ = & p^{3}+q^{3}+3 p^{2} q+3 p q^{2}+p^{3}+q^{3}+3 p^{2} q \omega \\ & \left.+3 p \omega^{2} q^{2}+p^{3}+q^{3}+3 p^{2} q \omega^{2}+3 p q^{2} \omega\right) \\ = & 3\left(p^{3}+q^{3}\right)+3 p^{2} q\left(1+\omega+\omega^{2}\right)+ \\ \quad & 3 p q^{2}\left(1+\omega+\omega^{2}\right) \\ \therefore & x^{3}+y^{3}+z^{3}=3\left(p^{3}+q^{3}\right) \end{aligned}

Omega বিষয়ক টপিকের ওপরে পরীক্ষা দাও