Question
If $${x^3} - 1 = 0$$ has the non-real complex roots $$\alpha ,\beta $$ then the value of $${\left( {1 + 2\alpha + \beta } \right)^3} - {\left( {3 + 3\alpha + 5\beta } \right)^3}$$ is
A.
$$- 7$$
B.
$$6$$
C.
$$- 5$$
D.
$$0$$
Answer :
$$- 7$$
Solution :
$$\eqalign{
& {\text{Here given }}{x^3} - 1 = 0 \cr
& {\text{has roots }}\alpha \,\& \,\beta {\text{ which are complex}} \cr
& x = \alpha \,\& \,\alpha = \omega \cr
& x = \beta \,\& \,\beta = {\omega ^2} \cr
& {\text{Now, }}{\left( {1 + 2\alpha + \beta } \right)^3} - {\left( {3 + 3\alpha + 5\beta } \right)^3} \cr
& = {\left( {1 + 2\omega + {\omega ^2}} \right)^3} - {\left( {3 + 3\omega + 5{\omega ^2}} \right)^3} \cr
& = {\left( {1 + \omega + {\omega ^2} + \omega } \right)^3} - {\left( {3\left( {1 + \omega + {\omega ^2}} \right) + 2{\omega ^2}} \right)^3} \cr
& {\text{As we know, }}1 + \omega + {\omega ^2} = 0 \cr
& = {\left( {0 + \omega } \right)^3} - {\left( {3\left( 0 \right) + 2{\omega ^2}} \right)^3} \cr
& = {\omega ^3} - 8{\omega ^6} \cr
& = 1 - 8\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\omega ^3} = 1} \right) \cr
& = - 7 \cr
& {\text{So, }}{\left( {1 + 2\alpha + \beta } \right)^3} - {\left( {3 + 3\alpha + 5\beta } \right)^3} = - 7 \cr} $$