Exercise 2.2 Class 10 – Cube Roots Simplified
Question 1(part i, part ii, part iii, part iv ) Question 2(part i, part ii, part iii, part iv, part v, part vi, part vii, part viii) Question 3, Question 4, Question 5.
This post of “Exercise 2.2 class 10” delves into the topic of the cube roots of unity, a fundamental concept in complex numbers. In this exercise, you will explore the three distinct cube roots of unity, denoted as 1,ω, and \(\omega^2\), where ω is a complex number with special properties. Understanding the properties of ω and how it interacts with the cube roots of unity will enhance your problem solving skills and deepen your knowledge of complex numbers.
Question 1 Exercise 2.2 Class 10. Find the cube roots of -1, 8, -27, 64.
(i) Cube roots of -1
\begin{align*}
& \text{Let } x = (-1)^{\frac{1}{3}} \\
& x^3 = -1 \\
& x^3 + 1 = 0 \\
& (x^3 + 1) = 0 \\
& \therefore (a^3 + b^3) = (a + b)(a^2 – ab + b^2) \\
& (x + 1) \left[ x^2 – (x)(1) + 1^2 \right] = 0 \\
& (x + 1)(x^2 – x + 1) = 0 \\
& x + 1 = 0 \text{ or } x^2 – x + 1 = 0 \\
& x = -1 \text{ or } x^2 – x + 1 = 0 \\
& \text{Now we solve } x^2 – x + 1 = 0 \text{ by formula} \\
& ax^2 + bx + c = 0 \\
& a = 1, \, b = -1, \, c = 1 \\
& x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\
& = \frac{-(-1) \pm \sqrt{(-1)^2 – 4(1)(1)}}{2 \times 1} \\
& = \frac{1 \pm \sqrt{1 – 4}}{2} \\
& = \frac{1 \pm \sqrt{-3}}{2} \\
& = \frac{1 \pm \sqrt{-1 \times 3}}{2} \\
& = \frac{1 \pm i \sqrt{3}}{2} \\
& x = \frac{1 + i \sqrt{3}}{2} \text{ or } x = \frac{1 – i \sqrt{3}}{2} \\
& x = \omega \text{ or } x = \omega^2 \\
& \text{So, cube roots of -1 are -1, } \omega \text{ and } \omega^2
\end{align*}
(ii) Cube roots of 8
\begin{align*}
& \text{Let } x = (8)^{\frac{1}{3}} \\
& x^3 = 8 \\
& x^3 – 8 = 0 \\
& x^3 – 2^3 = 0 \\
& \therefore (a^3 – b^3) = (a – b)(a^2 + ab + b^2) \\
& (x – 2) \left[ x^2 + (x)(2) + 2^2 \right] = 0 \\
& (x – 2)(x^2 + 2x + 4) = 0 \\
& x – 2 = 0 \text{ or } x^2 + 2x + 4 = 0 \\
& x = 2 \\
& \text{Now we solve } x^2 + 2x + 4 = 0 \text{ by formula} \\
& a = 1, \, b = 2, \, c = 4 \\
& x = \frac{-2 \pm \sqrt{b^2 – 4ac}}{2a} \\
& = \frac{-2 \pm \sqrt{(2)^2 – 4(1)(4)}}{2 \times 1} \\
& = \frac{-2 \pm \sqrt{4 – 16}}{2} \\
& = \frac{-2 \pm \sqrt{-12}}{2} \\
& = \frac{-2 \pm \sqrt{4 \times (-3)}}{2} \\
& = \frac{-2 \pm 2 \sqrt{-3}}{2} \\
& = -1 \pm \sqrt{-3} \\
& = -1 \pm i \sqrt{3} \\
& x = -1 \pm i \sqrt{3} \\
& \text{So the cube roots of 8 are } 2,2\omega, 2\omega^2
\end{align*}
(iii) Cube roots of -27
\begin{align*}
&\text{Let } x = (-27)^{\frac{1}{3}} \\
&x^3 = -27 \\
&x^3 + 27 = 0 \\
&x^3 + 3^3 = 0 \\
&\text{Since } (a^3 + b^3) = (a + b)(a^2 – ab + b^2) \\
&(x + 3) \left[ x^2 – (x)(3) + 3^2 \right] = 0 \\
&\text{Either } (x + 3)(x^2 – 3x + 9) = 0 \\
&x + 3 = 0 \quad \text{or} \quad x^2 – 3x + 9 = 0 \\
&\text{Now we solve } x^2 – 3x + 9 = 0 \text{ by the quadratic formula} \\
&a = 1, \quad b = -3, \quad c = 9 \\
&x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\
&x = \frac{-(-3) \pm \sqrt{(-3)^2 – 4(1)(9)}}{2(1)} \\
&x = \frac{3 \pm \sqrt{9 – 36}}{2} \\
&x = \frac{3 \pm \sqrt{-27}}{2} \\
&x = \frac{3 \pm \sqrt{(9)(-3)}}{2} \\
&x = \frac{3 \pm 3\sqrt{-3}}{2} \\
&x = -3 \left( \frac{1 \pm i\sqrt{3}}{2} \right) \\
&x = -3 \left( \frac{-1 + i\sqrt{3}}{2} \right) \quad \text{or} \quad x = -3 \left( \frac{-1 – i\sqrt{3}}{2} \right) \\
&x = -3 \omega \quad \text{or} \quad x = -3\omega^2 \\
&\text{So cube roots of } -27 \text{ are } -3, -3\omega, \text{ and } -3\omega^2
\end{align*}
(iv) Cube roots of 64
\begin{align*}
&\text{Let } x = (64)^{\frac{1}{3}} \\
&x^3 = 64 \\
&x^3 – 64 = 0 \\
&x^3 – 4^3 = 0 \\
&\text{Since } a^3 – b^3 = (a – b)(a^2 + ab + b^2) \\
&(x – 4) \left[ x^2 + (x)(4) + 4^2 \right] = 0 \\
&(x – 4)(x^2 + 4x + 16) = 0 \\
&\text{Either } x – 4 = 0 \quad \text{or} \quad x^2 + 4x + 16 = 0 \\
&x = 4 \\
&\text{Now we solve } x^2 + 4x + 16 = 0 \text{ by the quadratic formula} \\
&a = 1, \quad b = 4, \quad c = 16 \\
&x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\
&x = \frac{-4 \pm \sqrt{4^2 – 4 \times 1 \times 16}}{2(1)} \\
&x = \frac{-4 \pm \sqrt{16 – 64}}{2} \\
&x = \frac{-4 \pm \sqrt{-48}}{2} \\
&x = \frac{-4 \pm \sqrt{16(-3)}}{2} \\
&x = \frac{-4 \pm 4\sqrt{-3}}{2} \\
&x = 4 \left( \frac{-1 \pm i\sqrt{3}}{2} \right) \\
&\text{Either } x = 4 \left( \frac{-1 + i\sqrt{3}}{2} \right) \quad \text{or} \quad x = 4 \left( \frac{-1 – i\sqrt{3}}{2} \right) \\
&\omega = \frac{-1 + i\sqrt{3}}{2}, \quad \omega^2 = \frac{-1 – i\sqrt{3}}{2} \\
&\text{Therefore, } x = 4\omega \quad \text{or} \quad x = 4\omega^2 \\
&\text{So cube roots of } 64 \text{ are } 4, 4\omega, \text{ and } 4\omega^2
\end{align*}
Question 2 Exercise 2.2 Class 10. Evaluate
(i) \((1 – \omega – \omega^2)^7\)
\begin{align*}
\text{Solution:} \quad (1 – \omega – \omega^2)^7 \\
= & \left[ 1 – (\omega + \omega^2) \right]^7 \\
= & \left[ 1 – (-1) \right]^7 \quad \text{(since } \omega + \omega^2 = -1 \text{)} \\
= & (1 + 1)^7 \\
= & 2^7 \\
= & 128
\end{align*}
(ii) \((1 – 3\omega – 3\omega^2)^5\)
\begin{align*}
\text{Solution:} \quad (1 – 3\omega – 3\omega^2)^5 \\
= & \left[ 1 – 3(\omega + \omega^2) \right]^5 \\
= & \left[ 1 – 3(-1) \right]^5 \quad \text{(since } \omega + \omega^2 = -1 \text{)} \\
= & (1 + 3)^5 \\
= & 4^5 \\
= & 1024
\end{align*}
(iii)\((9 + 4\omega + 4\omega^2)^3\)
\begin{align*}
\text{Solution:} \quad (9 + 4\omega + 4\omega^2)^3 \\
= & \left[ 9 + 4(\omega + \omega^2) \right]^3 \\
= & \left[ 9 + 4(-1) \right]^3 \quad \text{(since } \omega + \omega^2 = -1 \text{)} \\
= & (9 – 4)^3 \\
= & 5^3 \\
= & 125
\end{align*}
(iv)\((2 + 2\omega – 2\omega^2)(3 – 3\omega + 3\omega^2)\)
\begin{align*}
\text{Solution:} \quad (2 + 2\omega – 2\omega^2)(3 – 3\omega + 3\omega^2) \\
= & \left[ 2(1 + \omega – \omega^2) \right] \left[ 3(1 + \omega – \omega^2) \right] \\
= & \left[ 2(1 + \omega – \omega^2) \right] \left[ 3(1 + \omega^2 – \omega) \right] \\
\text{Since } 1 + \omega + \omega^2 = 0: \\
1 + \omega = & -\omega^2 \quad \text{and} \quad 1 + \omega^2 = -\omega \\
= & [2(-\omega^2 – 2\omega)][3(-\omega – 3\omega)] \\
= & (-4\omega^2 – 6\omega) \\
= & 24(\omega^3) \\
= & 24(1) \quad \text{(since } \omega^3 = 1 \text{)} \\
= & 24
\end{align*}
(v) \(\left( -1 + \sqrt{3} \right)^6 + \left( -1 – \sqrt{3} \right)^6\)
\begin{align*}
\text{Solution:} \quad \left( -1 + \sqrt{3} \right)^6 + \left( -1 – \sqrt{3} \right)^6 \quad \dots (i) \\
\text{As } \frac{-1 + \sqrt{3}}{2} = \omega \quad \text{and} \quad \frac{-1 – \sqrt{3}}{2} = \omega^2 \\
-1 + \sqrt{3} = & 2\omega \quad \text{and} \quad -1 – \sqrt{3} = 2\omega^2 \\
\text{Now equation } (i) \text{ becomes:} \\
= & \left( 2\omega \right)^6 + \left( 2\omega^2 \right)^6 \\
= & 2^6\omega^6 + 2^6\omega^{12} \\
= & 2^6 \left( \left( \omega^3 \right)^2 + \left( \omega^3 \right)^4 \right) \quad \text{(as } \omega^3 = 1 \text{)} \\
= & 2^6 \left( 1^2 + 1^4 \right) \\
= & 64(1 + 1) \\
= & 64(2) \\
= & 128
\end{align*}
(vi) \(\frac{-1 + \sqrt{3}}{2}^9 + \frac{-1 – \sqrt{3}}{2}^9\)
\begin{align*}
\text{Solution:} \quad \frac{-1 + \sqrt{3}}{2}^9 + \frac{-1 – \sqrt{3}}{2}^9 \quad \dots (i) \\
\text{As } \frac{-1 + \sqrt{3}}{2} = \omega \quad \text{and} \quad \frac{-1 – \sqrt{3}}{2} = \omega^2 \\
\text{Now equation } (i) \text{ becomes:} \\
= & \omega^9 + \left( \omega^2 \right)^9 \\
= & \omega^9 + \omega^8 \\
= & \left( \omega^3 \right)^3 + \left( \omega^3 \right)^6 \quad \text{(as } \omega^3 = 1 \text{)} \\
= & 1 + 1 \\
= & 2
\end{align*}
(vii)\(\omega^7 + \omega^{38} – 5\)
\begin{align*}
\text{Solution:} \quad \omega^7 + \omega^{38} – 5 \\
= & \omega^7 \omega^0 + \omega^6 \omega^5 – 5 \\
= & \left( \omega^3 \right)^12 \omega^2 – 5 \\
= & (1)^12 \omega^2 – 5 \quad \text{(as } \omega^3 = 1 \text{)} \\
= & 1 + 1\omega^2 \quad – 5 \\
= & \left( \omega^7 -5 \right) \\
= & -1- \omega^6
\end{align*}
(viii) \(\omega^{-13} + \omega^{-17}\)
\begin{align*}
\omega^{-13} + \omega^{-17} & = \frac{1}{\omega^{13}} + \frac{1}{\omega^{17}} \\
& = \frac{1}{\omega^{12} \omega} + \frac{1}{\omega^{15} \omega^2} \\
& = \frac{1}{(\omega^3)^4 \omega} + \frac{1}{(\omega^3)^5 \omega^2} \\
& = \frac{1}{(1)^4 \omega} + \frac{1}{(1)^5 \omega^2} \quad \because \omega^3 = 1 \\
& = \frac{1}{\omega} + \frac{1}{\omega^2} \\
& = \frac{\omega^2 + \omega}{\omega (\omega^2)} \\
& = \frac{-1}{\omega^3} \quad \because 1 + \omega + \omega^2 = 0 \\
& = \frac{-1}{1} \\
& = -1
\end{align*}
Question 3 Exercise 2.2 Class 10.
\( \text{Prove that,} \quad x^3 + y^3= (x + y)(x + \omega y)(x + \omega^2 y) \)
\(\text{Let,} \quad \text{R.H.S.}= (x + y)(x + \omega y)(x + \omega^2 y)\)
\begin{align*}
&= (x + y)(x^2 + \omega^2 xy + \omega xy + \omega^3 y^2) \\
&= (x + y) \left[ x^2 + (\omega^2 + \omega)xy + \omega^3 y^2 \right] \\
&\because 1 + \omega + \omega^2 = 0 \implies \omega + \omega^2 = -1 \quad \text{and} \quad \omega^3 = 1 \\
&= (x + y) \left( x^2 + (-1)xy + y^2 \right) \\
&= (x + y)(x^2 – xy + y^2) \\
\\
\text{Using the formula:} \quad a^3 + b^3 &= (a + b)(a^2 – ab + b^2) \\
\\
\therefore x^3 + y^3 &= \text{L.H.S.}
\end{align*}
Question 4 of Exercise 2.2 Class 10. Prove that
\(\quad x^3 + y^3 + z^3 – 3xyz = (x + y + z)(x + \omega y + \omega^2 z)(x + \omega^2 y + \omega z) \)
\(\text{Let R.H.S.} = (x + y + z)(x + \omega y + \omega^2 z)(x + \omega^2 y + \omega z)\)
\( = (x + y + z)(x^2 + \omega^2 xy + \omega xz + \omega yx + \omega^3 y^2 + \omega^2 y^2 z + \omega^4 z^2) \)
\( = (x + y + z) \left[ x^2 + \omega^3 y^2 + \omega^3 z^2 + (\omega^2 + \omega)xy + (\omega^2 + \omega^4)yz + (\omega + \omega^2)zx \right] \)
\( \because 1 + \omega + \omega^2 = 0 \implies \omega + \omega^2 = -1 \quad \text{and} \quad \omega^3 = 1 \)
\( = (x + y + z) \left( x^2 + y^2 + z^2 – xy – yz – zx \right) \)
\( = x^3 + y^3 + z^3 – 3xyz = \text{L.H.S.} \)
\(\text{Using the formula:} \quad(a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca) = a^3 + b^3 + c^3 – 3abc \)
Question 5 of Exercise 2.2 Class 10. Prove that
\(\quad (1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \dots= 1 \quad \text{for 2n factors} \)
\(\text{Let,} \quad \text{L.H.S.}=(1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \dots 2n \text{ factors}\)
\begin{align*}
\\
&= (1 + \omega)(1 + \omega^2)(1 + \omega^3)(1 + \omega^6) \dots 2n \text{ factors} \\
&\because \omega^3 = 1 \implies \omega^6 = (\omega^3)^2 = 1 \\
&= (1 + \omega)(1 + \omega^2)(1 + \omega)(1 + \omega^2) \dots n \text{ factors} \\
&= \left[ (1 + \omega)(1 + \omega^2) \right]^n \\
&= \left[ 1 + \omega^2 + \omega + \omega^3 \right]^n \\
&= \left[ 1 + \omega + \omega^2 + \omega^3 \right]^n \\
&= \left[ 0 + 1 \right]^n \\
&= 1^n \\
&= 1 = \text{R.H.S.} \quad \therefore \text{L.H.S.} = \text{R.H.S.}
\end{align*}
We hope this solution of exercise 2.2 class 10 helps you in your preparation of exams. If you need a refresher on previous concepts, take a look at Exercise 2.1 on finding the discriminant and the nature of roots. Now let’s move on to Exercise 2.3. For an in-depth understanding of cube roots of unity and their applications, check out this informative Cuemath article on Cube Roots of Unity.
Here’s a video lecture of exercise 2.2 class 10 to further enhance you concepts.