If w is the complex cube root of unity, then is \[{{\left( 2+5w+2{{w}^{2}} \right)}^{6}}\] = \[{{\left( 2+2w+5{{w}^{2}} \right)}^{6}}\] = 729?
(a) True
(b) False
Answer
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Hint: To solve this problem, we should know the basics of complex numbers and their properties. In this problem, we should know about the following properties of the cube root of unity. These properties are: ${{w}^{3}}=1$ (by definition)
$1+w+{{w}^{2}}=0$
We will use these properties to solve this problem.
Complete step-by-step answer:
For solving the questions, we should know about the basic properties of the cube root of unity and how we can manipulate the terms in the expression to find the answer. By definition, the cube root of unity means the cube root of 1. These roots are $w,{{w}^{2}},{{w}^{3}}(=1)$. The value of w is $\dfrac{-1+\sqrt{3}}{2}$. Also, another property of cube roots of unity is-
$1+w+{{w}^{2}}$ = 0 -- (1)
Thus, coming back to the problem in hand, we first start by evaluating \[{{\left( 2+5w+2{{w}^{2}} \right)}^{6}}\], we have,
= \[{{\left( 2+5w+2{{w}^{2}} \right)}^{6}}\]
= \[{{\left( 2(1+{{w}^{2}})+5w \right)}^{6}}\]
From property (1), we have,
= \[{{\left( 2(-w)+5w \right)}^{6}}\]
= ${{(3w)}^{6}}$
= 729 ${{\left( {{w}^{3}} \right)}^{2}}$
Also, since, ${{w}^{3}}=1$. Thus, we have,
= 729 -- (A)
Now, we evaluate \[{{\left( 2+2w+5{{w}^{2}} \right)}^{6}}\], we have,
= \[{{\left( 2+2w+5{{w}^{2}} \right)}^{6}}\]
= \[{{\left( 2(1+w)+5{{w}^{2}} \right)}^{6}}\]
From property (1), we have,
= \[{{\left( 2(-{{w}^{2}})+5{{w}^{2}} \right)}^{6}}\]
= ${{(3{{w}^{2}})}^{6}}$
= 729 ${{\left( {{w}^{3}} \right)}^{4}}$
Also, since, ${{w}^{3}}=1$. Thus, we have,
= 729 -- (B)
Thus, from (A) and (B), we can conclude that \[{{\left( 2+5w+2{{w}^{2}} \right)}^{6}}\] = \[{{\left( 2+2w+5{{w}^{2}} \right)}^{6}}\] = 729.
Hence, the correct answer is (a) True.
Note: It is important to know the basic properties of the cube root of unity. To extend this, one can also learn the properties of ${{n}^{th}}$ root of unity, which will give a general formula for any root of unity in general. This basically means finding the roots of ${{1}^{\dfrac{1}{n}}}$. Most of the results are analogous to the cube roots of unity and these results can be useful to solve more complex problems.
$1+w+{{w}^{2}}=0$
We will use these properties to solve this problem.
Complete step-by-step answer:
For solving the questions, we should know about the basic properties of the cube root of unity and how we can manipulate the terms in the expression to find the answer. By definition, the cube root of unity means the cube root of 1. These roots are $w,{{w}^{2}},{{w}^{3}}(=1)$. The value of w is $\dfrac{-1+\sqrt{3}}{2}$. Also, another property of cube roots of unity is-
$1+w+{{w}^{2}}$ = 0 -- (1)
Thus, coming back to the problem in hand, we first start by evaluating \[{{\left( 2+5w+2{{w}^{2}} \right)}^{6}}\], we have,
= \[{{\left( 2+5w+2{{w}^{2}} \right)}^{6}}\]
= \[{{\left( 2(1+{{w}^{2}})+5w \right)}^{6}}\]
From property (1), we have,
= \[{{\left( 2(-w)+5w \right)}^{6}}\]
= ${{(3w)}^{6}}$
= 729 ${{\left( {{w}^{3}} \right)}^{2}}$
Also, since, ${{w}^{3}}=1$. Thus, we have,
= 729 -- (A)
Now, we evaluate \[{{\left( 2+2w+5{{w}^{2}} \right)}^{6}}\], we have,
= \[{{\left( 2+2w+5{{w}^{2}} \right)}^{6}}\]
= \[{{\left( 2(1+w)+5{{w}^{2}} \right)}^{6}}\]
From property (1), we have,
= \[{{\left( 2(-{{w}^{2}})+5{{w}^{2}} \right)}^{6}}\]
= ${{(3{{w}^{2}})}^{6}}$
= 729 ${{\left( {{w}^{3}} \right)}^{4}}$
Also, since, ${{w}^{3}}=1$. Thus, we have,
= 729 -- (B)
Thus, from (A) and (B), we can conclude that \[{{\left( 2+5w+2{{w}^{2}} \right)}^{6}}\] = \[{{\left( 2+2w+5{{w}^{2}} \right)}^{6}}\] = 729.
Hence, the correct answer is (a) True.
Note: It is important to know the basic properties of the cube root of unity. To extend this, one can also learn the properties of ${{n}^{th}}$ root of unity, which will give a general formula for any root of unity in general. This basically means finding the roots of ${{1}^{\dfrac{1}{n}}}$. Most of the results are analogous to the cube roots of unity and these results can be useful to solve more complex problems.
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