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Chapter 2: Problem 139

If cube root of unity are \(1, w, w^{2}\) then the roots of the equation \((\mathrm{x}-1)^{3}+8=0\) are (a) \(-1,-1,-1\) (b) \(-1,-1+2 \mathrm{w},-1-2 \mathrm{w}^{2}\) (c) \(-1,1+2 \mathrm{w}, 1+2 \mathrm{w}^{2}\) (d) \(-1,1-2 \mathrm{w},+1-2 \mathrm{w}^{2}\)

Short Answer

Expert verified
The short answer to the question is: (d) \(-1, 1 - 2w, 1 - 2w^2\).

Step by step solution

01

Rewrite the given equation

Rewrite the given equation \((x-1)^3 + 8 = 0\) as \[(x-1)^3 = -8\]
02

Express -8 as -2 times cube root of unity

Knowing that \(w^3 = 1\), we can express -8 as a product of -2 and cube root of unity: \[-8 = -2 \cdot (1 + w + w^2)\]
03

Make x the subject

Now, we can write x in terms of the cube roots of unity: \[(x-1)^3 = -2(1 + w + w^2) \Rightarrow x = 1 - \sqrt[3]{2(1 + w + w^2)}\]
04

Find the three roots

Substitute the cube roots of unity \(1, w, w^2\) into the expression for x: 1. When the cube root of unity is 1: \[x_1 = 1 - \sqrt[3]{2(1 + 1 + 1)} = 1 - 2 = -1\] 2. When the cube root of unity is w: \[x_2 = 1 - \sqrt[3]{2(1 + w + w^2)} = 1 - 2w\] 3. When the cube root of unity is \(w^2\): \[x_3 = 1 - \sqrt[3]{2(1 + w^2 + w^4)} = 1 - 2w^2\]
05

Compare the roots with the given options

Comparing the three roots we found, \(x_1 = -1\), \(x_2 = 1 - 2w\), and \(x_3 = 1 - 2w^2\), with the given options, we can see that the correct answer is: (d) \(-1, 1 - 2w, 1 - 2w^2\)

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Most popular questions from this chapter

If \(Z=\cos (4 \pi / 7)+i \sin (4 \pi / 7)\) then \(\operatorname{Re}\left(z+z^{2}+z^{3}\right)=\) (a) \(\cos (\pi / 3)\) (b) \(\cos (2 \pi / 3)\) (c) \(\cos (\pi / 6)\) (d) \(\cos (5 \pi / 6)\)

\(\left[\left\\{(\cos 2 \theta-\mathrm{i} \sin 2 \theta)^{7}(\cos 3 \theta+\mathrm{i} \sin 3 \theta)^{-5}\right\\}\right.\) \(/\left\\{(\cos 4 \theta+i \sin 4 \theta)^{12}(\cos 5 \theta+i \sin 5 \theta)^{-6}\right]=\) (a) \(\cos 33 \theta+i \sin 33 \theta\) (b) \(\cos 33 \theta-\mathrm{i} \sin 33 \theta\) (c) \(\cos 47 \theta+i \sin 47 \theta\) (d) \(\cos 47 \theta+\mathrm{i} \sin 47 \theta\)

If \(\mathrm{a}=\cos (2 \pi / 7)+\mathrm{i} \sin (2 \pi / 7)\) then the quadratic equation whose roots are \(\alpha=\mathrm{a}+\mathrm{a}^{2}+\mathrm{a}^{4}\) and \(\beta=\mathrm{a}^{3}+\mathrm{a}^{5}+\mathrm{a}^{6}\) is (a) \(x^{2}-x+2\) (b) \(x^{2}+x-2\) (c) \(x^{2}-x-2\) (d) \(x^{2}+x+2\)

The complex numbers \(\sin x+i \cos 2 x\) and \(\cos x-i \sin 2 x\) are conjugate of each other for (a) \(\mathrm{x}=\mathrm{k} \pi, \mathrm{k} \in \mathrm{z}\) (b) \(x=0\) (c) \(\mathrm{x}=[\mathrm{k}+(1 / 2)] \pi, \mathrm{k} \in \mathrm{z}\) (d) no value of \(\mathrm{x}\)

If \(z=\cos (\pi / 3)-i \sin (\pi / 3)\) then \(z^{2}-z+1=\) (a) \(-2 \mathrm{i}\) (b) 2 (c) 0 (d) \(-2\)
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