Official PYQ · 1995Q2.8Calculus: Limits, Continuity, DifferentiabilityMCQ1 mark

GATE CS 1995 Q2.8 — Calculus: Limits, Continuity, Differentiability

If the cube roots of unity are $1, \omega$ and $\omega^2$ , then the roots of the following equation are $(x-1)^3 +8 =0$

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Answer: C

We are given the equation:

(x-1)^3 + 8 = 0

Rearranging the terms, we get:

(x-1)^3 = -8

Taking the cube root on both sides:

x-1 = (-8)^{1/3}

Since $(-8)^{1/3} = -2 \times (1)^{1/3}$ , and the cube roots of unity are $1, \omega, \text{ and } \omega^2$ , the three possible values for $x-1$ are:

x-1 = -2(1) = -2

x-1 = -2\omega

x-1 = -2\omega^2

Solving for $x$ in each case:

$x = 1 - 2 = -1$
$x = 1 - 2\omega$
$x = 1 - 2\omega^2$

Thus, the roots of the equation are $-1, 1 - 2\omega, \text{ and } 1 - 2\omega^2$ .

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Published 10 Oct 2026

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