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Chapter 2: Problem 126

If \(1, \mathrm{w}\) and \(\mathrm{w}^{2}\) are cube root of 1 then \((1-\mathrm{w})\left(1-\mathrm{w}^{2}\right)\) \(\left(1-\mathrm{w}^{4}\right)\left(1-\mathrm{w}^{8}\right)=\) (a) 16 (b) 8 (c) 9 (d) 64

Short Answer

Expert verified
The simplified expression is \(2(3 - 4w + w^2)\), which evaluates to -2. However, this value is not in any of the provided answer choices. There might be an error in the problem or the answer choices.

Step by step solution

01

Recall the cube roots of 1

Because w and \(w^2\) are cube roots of 1, we can write \(w^3 =1\) and \((w^2)^3={w^6}=1\).
02

Find the values of \(w^4\) and \(w^8\)

Since \(w^3 = 1\), we can use it to determine the values of \(w^4\) and \(w^8\). To find the value of \(w^4\), notice that \(w^4 = w\cdot w^3 = w(1)=w\). Then, finding the value of \(w^8\) can be done by realizing that \(w^8 = w^5\cdot w^3 = (w^2)(1)= w^2\).
03

Simplify the given expression

Now, we can substitute the values of \(w^4\) and \(w^8\) into the given expression: \((1-w)(1-w^2)(1-w^4)(1-w^8) = (1-w)(1-w^2)(1-w)(1-w^2)\) Notice that we have the same term repeated twice. We can rewrite the expression as: \((1-w)^2(1-w^2)^2\)
04

Expand and simplify the expression

Expand and simplify the expression as follows: \((1-w)^2(1-w^2)^2 = [(1-2w+w^2)(1-2w^2+w^4)]\) Now let's substitute the values for \(w^4 = w\) and \(w^6 = 1\) to simplify the expression: \([(1-2w+w^2)(1-2w^2+w^6w^2)] = [(1-2w+w^2)(1-2w^2+w^2)]\) Now, expand the expression: \((1-2w+w^2)(1-2w^2+w^2) = 1 - 2w^2 + w^2 - 2w + 4w^3 - 2w^4 + w^2 - 2w^4 + w^6\) We know \(w^3 = 1\) and \(w^6 = 1\), so we can substitute these values: \(1 - 2w^2 + w^2 - 2w + 4w^3 - 2w^4 + w^2 - 2w^4 + 1 = 2 - 4w + 2w^2 + 4 - 4w = 6 - 8w +2w^2\)
05

Factor the expression and evaluate

Now, factor the simplified expression: \(6 - 8w + 2w^2 = 2(3 - 4w + w^2)\) Substitute values for w using the fact that \(w^3=1\): \(2(3 - 4(1) + 1^2) = 2(-1) = -2\) Now, since this value is not in any of the options given (a, b, c, d), we can conclude that there is likely an error in the problem or in the answer choices provided.

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Most popular questions from this chapter

The small positive integer ' \(n\) ' for which \((1+i)^{2 n}=(1-i)^{2 n}\) is \(\begin{array}{lll}\text { (a) } 4 & \text { (b) } 8 & \text { (c) } 2\end{array}\) (d) 12

Let \(z=[(3+2 i \sin \theta) /(1-2 i \sin \theta)]\) and \(z=\underline{z}\) then \(\theta=\) (a) \((2 \mathrm{k}+1)(\pi / 2), \mathrm{k} \in \mathrm{z}\) (b) \(2 \mathrm{k} \pi, \mathrm{k} \overline{\in \mathrm{z}}\) (c) \(k \pi, k \in z\) (d) None

If \(\mathrm{x}>1\) then the square roots of \(1-\sqrt{\left(\mathrm{x}^{2}-1\right) \mathrm{i}}\) (a) \(\pm[\sqrt{\\{}(\mathrm{x}+1) / 2\\}-\mathrm{i} \sqrt{\\{}(\mathrm{x}-1) / 2\\}]\) (b) \(\pm[\sqrt{\\{}(\mathrm{x}+1) / 2\\}+\mathrm{i} \sqrt{\\{}(\mathrm{x}-1) / 2\\}]\) (c) \(\pm[\sqrt{\\{}(\mathrm{x}-1) / 2\\}-\mathrm{i} \sqrt{\\{}(\mathrm{x}+1) / 2\\}]\) (d) \(\pm[\sqrt{\\{}(\mathrm{x}-1) / 2\\}+\mathrm{i} \sqrt{\\{}(\mathrm{x}+1) / 2\\}]\)

If \(z=\cos \theta+i \sin \theta\) then \(\arg \left(z^{2}+\underline{z}\right)=\) (a) \((3 \theta / 2)\) (b) \(\theta\) (c) \((\theta / 2)\) (d) \(3 \theta\)

Let \(z\) be complex number with modulus 2 and argument \([(-2 \pi) / 3]\) then \(\mathrm{z}=\) (a) \(-1+\mathrm{i} \sqrt{3}\) (b) \([(-1+i \sqrt{3}) / 2]\) (c) \(-1-\mathrm{i} \sqrt{3}\) (d) \([(-1-i \sqrt{3}) / 2]\)
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