`(-1/2 + sqrt(3)/2 i )^3` Use the Binomial Theorem to expand the complex number, then simplify the result.

The expansion of `(x+y)^n` is,

`_nC_0x^n+_nC_1x^(n-1)y^1+_nC_2x^(n-2)y^2+.......+_nC_ny^n`

`=x^n+nx^(n-1)y+(n!)/(2!(n-2)!)x^(n-2)y^2+......+y^n`

`:.(-1/2+sqrt(3)/2i)^3)=(-1/2)^3+3(-1/2)^(3-1)*(sqrt(3)/2i)+_3C_2*(-1/2)^(3-2)*(sqrt(3)/2i)^2+(sqrt(3)/2i)^3`

`=-1/8+3(-1/2)^2*(sqrt(3)/2i)+(3!)/(2!(3-2)!)*(-1/2)^1*(sqrt(3)/2i)^2+(sqrt(3)/2i)^3`

`=-1/8+((3sqrt(3))/8)i-3/2*((3i^2)/4)+(3sqrt(3)i^3)/8`

`=-1/8+(3sqrt(3)i)/8-9/8i^2+(3sqrt(3)i^3)/8`

plug in `i^2=-1 `

`=-1/8+(3sqrt(3)i)/8-9/8(-1)+(3sqrt(3)(-i))/8`

`=-1/8+(3sqrt(3)i)/8+9/8-(3sqrt(3)i)/8`

`=-1/8+9/8`

`=(-1+9)/8`

`=8/8`

`=1`