First slide

Algebra of complex numbers

Question

Let $ω \neq 1,$ be a cube root of unity, and $a, b \in R$ .
Statement $- 1$ : $a^{3} + b^{3} = (a + b) (a ω + b ω^{2}) (a ω^{2} + b ω)$
Statement $- 2$ : $x^{3} - 1 = (x - 1) (x ω^{2} - ω) (x ω - ω^{2})$ for each $x \in R .$

Moderate

A

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
B

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
C

STATEMENT-1 is True, STATEMENT-2 is False
D

STATEMENT-1 is False, STATEMENT-2 is True

Solution

We have $- x$
$\begin{array}{l} (x ω^{2} - ω) (x ω - ω^{2}) \\ = x^{2} ω^{3} - x ω^{2} - x ω^{4} + ω^{3} \\ = x^{2} - (ω^{2} + ω) x + 1 \\ = x^{2} + x + 1 \end{array} [∵ ω^{2} + ω = - 1]$
$\begin{array}{l} ∴ (x - 1) (x ω^{2} - ω) (x ω - ω^{2}) \\ = (x - 1) (x^{2} + x + 1) = x^{3} - 1 \end{array}$
Thus, Statement $- 2$ is true. Replacing $x$ by –x, we get
$\begin{array}{l} (- x)^{3} - 1 = (- x - 1) (- x ω^{2} - ω) (- x ω - ω^{2}) \\ \Rightarrow x^{3} + 1 = (x + 1) (x ω^{2} + ω) (x ω + ω^{2}) \end{array}$
Taking conjugate of both the sides, we get
$x^{3} + 1 = (x + 1) (x ω + ω^{2}) (x ω^{2} + ω)$
$[∴ ω = ω^{2}]$
If $b = 0,$ , statement $- 1$ is clearly true. Suppose $b \neq 0 .$
Replacing $x$ by $a / b$ we get
$\begin{aligned} {(\frac{a}{b})}^{3} + 1 = (\frac{a}{b} + 1) (\frac{a}{b} w + w^{2}) (\frac{a}{b} w^{2} + w) \\ \Rightarrow a^{3} + b^{3} = (a + b) (a w + b w^{2}) (a w^{2} + b w) \end{aligned}$
Thus, statement $- 1$ is also true and statement $- 2$ is a correct explanation for it.

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