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Question: If 1, \[\omega \], \[{\omega ^2}\] are cube the roots of unity, then find the values of the followin...

If 1, ω\omega , ω2{\omega ^2} are cube the roots of unity, then find the values of the following.
i)(a+b)3+(aω+bω2)3+(aω2+bω)3{\left( {a + b} \right)^3} + {\left( {a\omega + b{\omega ^2}} \right)^3} + {\left( {a{\omega ^2} + b\omega } \right)^3}
ii)(a+2b)2+(aω2+2bω)2+(aω+2bω2)2{\left( {a + 2b} \right)^2} + {\left( {a{\omega ^2} + 2b\omega } \right)^2} + {\left( {a\omega + 2b{\omega ^2}} \right)^2}

Explanation

Solution

Here, we are required to find the values of the above parts when it is given that 1, ω\omega , ω2{\omega ^2} are cube the roots of unity. We know that the sum of three cube roots of unity is 0. We will expand the given expressions and take the like terms common and finally, substituting the fact that 1+ω+ω2=01 + \omega + {\omega ^2} = 0, will help us reach the required answer.

Formula Used:
We will use the following formulas:
1.(a+b)3=a3+3a2b+3ab2+b3{\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}
2.(a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}
3.1+ω+ω2=01 + \omega + {\omega ^2} = 0

Complete step-by-step answer:
According to the question, if 1, ω\omega , ω2{\omega ^2} are cube the roots of unity.
As sum of three cube roots of unity is 0, so 1+ω+ω2=01 + \omega + {\omega ^2} = 0
(a+b)3+(aω+bω2)3+(aω2+bω)3{\left( {a + b} \right)^3} + {\left( {a\omega + b{\omega ^2}} \right)^3} + {\left( {a{\omega ^2} + b\omega } \right)^3}
Now, using the formula (a+b)3=a3+3a2b+3ab2+b3{\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}, we get
Let (a+b)3+(aω+bω2)3+(aω2+bω)3=S{\left( {a + b} \right)^3} + {\left( {a\omega + b{\omega ^2}} \right)^3} + {\left( {a{\omega ^2} + b\omega } \right)^3} = S
S=a3+3a2b+3ab2+b3+a3+3a2bω+3ab2ω2+b3+a3+3a2bω2+3ab2ω+b3S = {a^3} + 3{a^2}b + 3a{b^2} + {b^3} + {a^3} + 3{a^2}b\omega + 3a{b^2}{\omega ^2} + {b^3} + {a^3} + 3{a^2}b{\omega ^2} + 3a{b^2}\omega + {b^3}
Taking the like terms common, we get,
⇒S=3a3+3a2b(1+ω+ω2)+3ab2(1+ω+ω2)+3b3\Rightarrow S = 3{a^3} + 3{a^2}b\left( {1 + \omega + {\omega ^2}} \right) + 3a{b^2}\left( {1 + \omega + {\omega ^2}} \right) + 3{b^3}
But,1+ω+ω2=01 + \omega + {\omega ^2} = 0, therefore
⇒S=3a3+3a2b(0)+3ab2(0)+3b3\Rightarrow S = 3{a^3} + 3{a^2}b\left( 0 \right) + 3a{b^2}\left( 0 \right) + 3{b^3}
⇒S=3a3+3b3=3(a3+b3)\Rightarrow S = 3{a^3} + 3{b^3} = 3\left( {{a^3} + {b^3}} \right)
Hence, the value of (a+b)3+(aω+bω2)3+(aω2+bω)3=3(a3+b3){\left( {a + b} \right)^3} + {\left( {a\omega + b{\omega ^2}} \right)^3} + {\left( {a{\omega ^2} + b\omega } \right)^3} = 3\left( {{a^3} + {b^3}} \right)
(a+2b)2+(aω2+2bω)2+(aω+2bω2)2{\left( {a + 2b} \right)^2} + {\left( {a{\omega ^2} + 2b\omega } \right)^2} + {\left( {a\omega + 2b{\omega ^2}} \right)^2}
Let (a+2b)2+(aω2+2bω)2+(aω+2bω2)2=T{\left( {a + 2b} \right)^2} + {\left( {a{\omega ^2} + 2b\omega } \right)^2} + {\left( {a\omega + 2b{\omega ^2}} \right)^2} = T
Using the formula (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}, we get
⇒T==a2+4ab+4b2+a2ω+4ab+4b2ω2+a2ω2+4ab+4b2ω\Rightarrow T = = {a^2} + 4ab + 4{b^2} + {a^2}\omega + 4ab + 4{b^2}{\omega ^2} + {a^2}{\omega ^2} + 4ab + 4{b^2}\omega
Taking the like terms common, we get,
⇒T=a2(1+ω+ω2)+12ab+4b2(1+ω+ω2)\Rightarrow T = {a^2}\left( {1 + \omega + {\omega ^2}} \right) + 12ab + 4{b^2}\left( {1 + \omega + {\omega ^2}} \right)
But, 1+ω+ω2=01 + \omega + {\omega ^2} = 0, hence,
⇒T=a2(0)+12ab+4b2(0)\Rightarrow T = {a^2}\left( 0 \right) + 12ab + 4{b^2}\left( 0 \right)
⇒T=12ab\Rightarrow T = 12ab
Hence, the value of (a+2b)2+(aω2+2bω)2+(aω+2bω2)2=12ab{\left( {a + 2b} \right)^2} + {\left( {a{\omega ^2} + 2b\omega } \right)^2} + {\left( {a\omega + 2b{\omega ^2}} \right)^2} = 12ab

Note: In this question we have assumed that 1+ω+ω2=01 + \omega + {\omega ^2} = 0. We can prove this equation as shown below.
Let us assume the cube root of unity or 1 as:
13=z\sqrt[3]{1} = z
Cubing both sides, we get
⇒1=z3\Rightarrow 1 = {z^3}
Or
⇒z3−1=0\Rightarrow {z^3} - 1 = 0
Now, using the formula (a3−b3)=(a−b)(a2+b2+ab)\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right), we get
⇒(z−1)(z2+z+1)=0\Rightarrow \left( {z - 1} \right)\left( {{z^2} + z + 1} \right) = 0
Therefore, either (z−1)=0\left( {z - 1} \right) = 0
⇒z=1\Rightarrow z = 1
Or, (z2+z+1)=0\left( {{z^2} + z + 1} \right) = 0
Comparing with (ax2+bx+c)=0\left( {a{x^2} + bx + c} \right) = 0
Here, a=1a = 1, b=1b = 1and c=1c = 1
Now, Determinant, D=b2−4acD = {b^2} - 4ac
Hence, for (z2+z+1)=0\left( {{z^2} + z + 1} \right) = 0,
D=(1)2−4×1=1−4=−3D = {\left( 1 \right)^2} - 4 \times 1 = 1 - 4 = - 3
Now, , Using quadratic formula,
z=−b±D2az = \dfrac{{ - b \pm \sqrt D }}{{2a}}
Here, a=1a = 1, b=1b = 1and c=1c = 1 and D=−3D = - 3
⇒z=−1±−32\Rightarrow z = \dfrac{{ - 1 \pm \sqrt { - 3} }}{2}
This can be written as:
⇒z=−1±3i2\Rightarrow z = \dfrac{{ - 1 \pm \sqrt 3 i}}{2}
Therefore, the three cube roots of unity are:
11, −12+3i2\dfrac{{ - 1}}{2} + \dfrac{{\sqrt 3 i}}{2} and −12−3i2\dfrac{{ - 1}}{2} - \dfrac{{\sqrt 3 i}}{2}
Now, according to the property, the sum of these three cube roots of unity will be equal to 0.
We know that,
1+ω+ω2=01 + \omega + {\omega ^2} = 0
Here, ω\omega represents the imaginary cube roots.
⇒1+−12+3i2+−12−3i2=1−1+0=0\Rightarrow 1 + \dfrac{{ - 1}}{2} + \dfrac{{\sqrt 3 i}}{2} + \dfrac{{ - 1}}{2} - \dfrac{{\sqrt 3 i}}{2} = 1 - 1 + 0 = 0
Hence, proved.