Question

If \[xyz\], \[y = (a\beta + b\alpha )\] and \[z = a\beta + b\alpha \] , and \[\alpha \] and another \[\beta \] are cube root of unity and not equal to 1 , then what is the value of xyz ?

Answer 1

Hint: In order to solve the question, you have to know that the cube root of unity is referred to as the Cube Root of 1. It is defined as the number that can be raised to the power of 3 and result is 1. suppose, \[\beta \] and \[\alpha \] are cube root of unity , then value of 1 + \[\alpha \] +\[\beta \]\[ = 0\] and value of \[1 \times \alpha \times \beta = 1\] . Here is the main key of this question, also if \[\alpha \] and \[\beta \] are cube root of unity then we can assume that
\[\beta \]=\[{\alpha ^2}\].\[{\alpha ^3} = 1\],\[{\beta ^3} = 1\]. Because alpha and beta are cube root of unity.

Complete answer:
First multiply \[xyz\], then move to others,
\[\; \Rightarrow xyz = (a + b)(a\alpha + b\beta ){w^n} = 1\]
 Then multiply each other’s, and go further.
$ \Rightarrow xyz = (a + b)({a^2}\alpha \beta + ab{\alpha ^2} + ab{\beta ^2} + {b^2}\alpha \beta )$
 Here, we will use properties of cube root of unity. Which I mentioned in hints,
$ \Rightarrow xyz = (a + b)({a^2}(1) + ab{\alpha ^2} + ab{\beta ^2} + {b^2}(1))$ ,
Because \[1 \times \alpha \times \beta = 1\]
 Now, take common ab, and after that put value of \[\beta = {\alpha ^2}\]
\[{\beta ^3} = 1\],\[{\alpha ^3} = 1\],
$ \Rightarrow xyz = (a + b)({a^2} + ab({\alpha ^2} + {({\alpha ^2})^2}) + {b^2})$
After further simplification,
$ \Rightarrow xyz = (a + b)({a^2} + ab({\alpha ^2} + {\alpha ^4}) + {b^2})$
$ \Rightarrow xyz = (a + b)({a^2} + ab({\alpha ^2} + {\alpha ^3}\alpha ) + {b^2})$
After some more further simplification, we will get,
$ \Rightarrow xyz = (a + b)({a^2} + ab({\alpha ^2} + (1)\alpha ) + {b^2})$
$ \Rightarrow xyz = (a + b)({a^2} + ab(( - 1)) + {b^2})$
Here \[1 + \alpha + {\alpha ^2} = 0\]because \[1 + \alpha + \beta = 0\] and \[\beta = {\alpha ^2}\]
$ \Rightarrow xyz = (a + b)({a^2} - ab + {b^2})$
And we will get bellowed equation,
$ \Rightarrow xyz = {a^3} + {b^3}$

Note:
Here for any power of root, remember two major formulas $1 + w + {w^2} + {w^3} + {w^4} + .... + {w^{n - 1}} = 0$ , where n is equal to 3,4,5…. According to the cube, the fourth, fifth root of unity. ${w^n} = 1$ Where n is equal to 3,4,5…. According to the cube, the fourth, fifth root of unity. Also use these ${a^3} + {b^3} = (a + b)({a^2} + {b^2} - ab)$ and ${a^3} - {b^3} = (a - b)({a^2} + {b^2} + ab)$, this is also useful. It is also called the de Mover system.