Exercise 2.2 Class 10 – Cube Roots Simplified
Exercise 2.2 Class 10 solutions are designed to help students master important concepts such as cube roots, the cube roots of unity (ω and ω²), and their applications in algebraic expressions. As a mathematics educator with years of experience guiding Class 10 students, I have prepared clear, step-by-step solutions that strictly follow the latest Punjab Board Class 10 Math textbook. All steps are presented using Unicode math expressions for easy reading on any device, helping students understand the logic behind each solution and strengthen their problem-solving skills for exams.
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.
Question 1 Exercise 2.2 Class 10. Find the cube roots of -1, 8, -27, 64.
(i) Cube roots of -1
Let x = (−1)^(1⁄3)
x³ = −1
x³ + 1 = 0
(a³ + b³) = (a + b)(a² − ab + b²)
(x + 1) [ x² − (x)(1) + 1² ] = 0
(x + 1)(x² − x + 1) = 0
x + 1 = 0 or x² − x + 1 = 0
x = −1 or x² − x + 1 = 0
Now we solve x² − x + 1 = 0 by formula
ax² + bx + c = 0
a = 1, b = −1, c = 1
\( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
\( = \frac{-(-1) \pm \sqrt{(-1)^2 – 4(1)(1)}}{2 \cdot 1} \)
\( = \frac{1 \pm \sqrt{1 – 4}}{2} \)
\( = \frac{1 \pm \sqrt{-3}}{2} \)
\( = \frac{1 \pm \sqrt{-1 \times 3}}{2} \)
\( = \frac{1 \pm i\sqrt{3}}{2} \)
\( x = \frac{1 + i\sqrt{3}}{2} \text{ or } \frac{1 – i\sqrt{3}}{2} \)
\( x = \omega \text{ or } \omega^2 \)
\( \text{So, cube roots of } -1 \text{ are } -1,\; \omega,\; \omega^2 \)
(ii) Cube roots of 8
Let x = 8^(1⁄3)
x³ = 8
x³ − 8 = 0
x³ − 2³ = 0
(a³ − b³) = (a − b)(a² + ab + b²)
(x − 2) [ x² + (x)(2) + 2² ] = 0
(x − 2)(x² + 2x + 4) = 0
x − 2 = 0 or x² + 2x + 4 = 0
x = 2
Now we solve x² + 2x + 4 = 0 by formula
a = 1, b = 2, c = 4
\( x = \frac{-2 \pm \sqrt{b^2 – 4ac}}{2a} \)
\( = \frac{-2 \pm \sqrt{(2)^2 – 4(1)(4)}}{2 \times 1} \)
\( = \frac{-2 \pm \sqrt{4 – 16}}{2} \)
\( = \frac{-2 \pm \sqrt{-12}}{2} \)
\( = \frac{-2 \pm \sqrt{4 \times (-3)}}{2} \)
\( = \frac{-2 \pm 2 \sqrt{-3}}{2} \)
\( = -1 \pm \sqrt{-3} \)
\( = -1 \pm i \sqrt{3} \)
\( x = -1 \pm i \sqrt{3} \)
\( \text{So the cube roots of 8 are } 2, 2\omega, 2\omega^2 \)
(iii) Cube roots of -27
Let x = (−27)⁽¹⁄³⁾
x³ = −27
x³ + 27 = 0
x³ + 3³ = 0
Since (a³ + b³) = (a + b)(a² − ab + b²)
(x + 3) [ x² − (x)(3) + 3² ] = 0
Either (x + 3)(x² − 3x + 9) = 0
x + 3 = 0 or x² − 3x + 9 = 0
Now we solve x² − 3x + 9 = 0 by the quadratic formula
a = 1, b = −3, c = 9
\( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
\( x = \frac{-(-3) \pm \sqrt{(-3)^2 – 4(1)(9)}}{2(1)} \)
\( x = \frac{3 \pm \sqrt{9 – 36}}{2} \)
\( x = \frac{3 \pm \sqrt{-27}}{2} \)
\( x = \frac{3 \pm \sqrt{(9)(-3)}}{2} \)
\( x = \frac{3 \pm 3\sqrt{-3}}{2} \)
\( x = -3 \left( \frac{1 \pm i\sqrt{3}}{2} \right) \)
\( x = -3 \left( \frac{-1 + i\sqrt{3}}{2} \right) \text{ or } x = -3 \left( \frac{-1 – i\sqrt{3}}{2} \right) \)
\( x = -3 \omega \text{ or } x = -3\omega^2 \)
\( \text{So cube roots of } -27 \text{ are } -3, -3\omega, \text{ and } -3\omega^2 \)
(iv) Cube roots of 64
Let x = 64⁽¹⁄³⁾
x³ = 64
x³ − 64 = 0
x³ − 4³ = 0
Since a³ − b³ = (a − b)(a² + ab + b²)
(x − 4) [ x² + (x)(4) + 4² ] = 0
(x − 4)(x² + 4x + 16) = 0
Either x − 4 = 0 or x² + 4x + 16 = 0
x = 4
Now we solve x² + 4x + 16 = 0 by the quadratic formula
a = 1, b = 4, c = 16
\( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
\( x = \frac{-4 \pm \sqrt{4^2 – 4 \times 1 \times 16}}{2(1)} \)
\( x = \frac{-4 \pm \sqrt{16 – 64}}{2} \)
\( x = \frac{-4 \pm \sqrt{-48}}{2} \)
\( x = \frac{-4 \pm \sqrt{16(-3)}}{2} \)
\( x = \frac{-4 \pm 4\sqrt{-3}}{2} \)
\( x = 4 \left( \frac{-1 \pm i\sqrt{3}}{2} \right) \)
\( \text{Either } x = 4 \left( \frac{-1 + i\sqrt{3}}{2} \right) \text{ or } x = 4 \left( \frac{-1 – i\sqrt{3}}{2} \right) \)
\( \omega = \frac{-1 + i\sqrt{3}}{2}, \quad \omega^2 = \frac{-1 – i\sqrt{3}}{2} \)
\( \text{Therefore, } x = 4\omega \text{ or } x = 4\omega^2 \)
\( \text{So cube roots of } 64 \text{ are } 4, 4\omega, \text{ and } 4\omega^2 \)
Question 2 Exercise 2.2 Class 10. Evaluate
(i) (1 − ω − ω²)⁷
Solution: (1 − ω − ω²)⁷
= [ 1 − (ω + ω²) ]⁷
= [ 1 − (−1) ]⁷ (since ω + ω² = −1)
= (1 + 1)⁷
= 2⁷
= 128
(ii) (1 − 3ω − 3ω²)⁵
Solution: (1 − 3ω − 3ω²)⁵
= [ 1 − 3(ω + ω²) ]⁵
= [ 1 − 3(−1) ]⁵ (since ω + ω² = −1)
= (1 + 3)⁵
= 4⁵
= 1024
(iii) (9 + 4ω + 4ω²)³
Solution: (9 + 4ω + 4ω²)³
= [ 9 + 4(ω + ω²) ]³
= [ 9 + 4(−1) ]³ (since ω + ω² = −1)
= (9 − 4)³
= 5³
= 125
(iv) (2 + 2ω − 2ω²)(3 − 3ω + 3ω²)
Solution: (2 + 2ω − 2ω²)(3 − 3ω + 3ω²)
= [ 2(1 + ω − ω²) ][ 3(1 + ω − ω²) ]
= [ 2(1 + ω − ω²) ][ 3(1 + ω² − ω) ]
Since 1 + ω + ω² = 0:
1 + ω = −ω² and 1 + ω² = −ω
= [ 2(−ω² − 2ω) ][ 3(−ω − 3ω) ]
= (−4ω² − 6ω)
= 24(ω³)
= 24(1) (since ω³ = 1)
= 24
(v) (−1 + √3)⁶ + (−1 − √3)⁶
Solution: (−1 + √3)⁶ + (−1 − √3)⁶ … (i)
As (−1 + √3)/2 = ω and (−1 − √3)/2 = ω²
−1 + √3 = 2ω and −1 − √3 = 2ω²
Now equation (i) becomes:
= (2ω)⁶ + (2ω²)⁶
= 2⁶ω⁶ + 2⁶ω¹²
= 2⁶ [ (ω³)² + (ω³)⁴ ] (as ω³ = 1)
= 2⁶ [ 1² + 1⁴ ]
= 64(1 + 1)
= 64(2)
= 128
(vi) ((−1 + √3)/2)⁹ + ((−1 − √3)/2)⁹
Solution: ((−1 + √3)/2)⁹ + ((−1 − √3)/2)⁹ … (i)
As (−1 + √3)/2 = ω and (−1 − √3)/2 = ω²
Now equation (i) becomes:
= ω⁹ + (ω²)⁹
= ω⁹ + ω⁸
= (ω³)³ + (ω³)⁶ (as ω³ = 1)
= 1 + 1
= 2
(vii) ω⁷ + ω³⁸ − 5
Solution: ω⁷ + ω³⁸ − 5
= ω⁷ ω⁰ + ω⁶ ω⁵ − 5
= (ω³)¹² ω² − 5
= 1¹² ω² − 5 (as ω³ = 1)
= 1 + 1ω² − 5
= (ω⁷ − 5)
= −1 − ω⁶
(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.
Prove that quad x³ + y³=(x + y)(x + ωy)(x + ω²y)
Let, R.H.S. = (x + y)(x + ωy)(x + ω²y)
= (x + y)(x² + ω²xy + ωxy + ω³y²)
= (x + y) [ x² + (ω² + ω)xy + ω³y² ]
∵ 1 + ω + ω² = 0 ⇒ ω + ω² = −1 and ω³ = 1
= (x + y) ( x² + (−1)xy + y² )
= (x + y)(x² − xy + y²)
Using the formula: a³ + b³ = (a + b)(a² − ab + b²)
∴ x³ + y³ = L.H.S.
Question 4 of Exercise 2.2 Class 10. Prove that
x³ + y³ + z³ − 3xyz = (x + y + z)(x + ωy + ω²z)(x + ω²y + ωz)
Let R.H.S. = (x + y + z)(x + ωy + ω²z)(x + ω²y + ωz)
= (x + y + z)(x² + ω²xy + ωxz + ωyx + ω³y² + ω²y²z + ω⁴z²)
= (x + y + z) [ x² + ω³y² + ω³z² + (ω² + ω)xy + (ω² + ω⁴)yz + (ω + ω²)zx ]
∵ 1 + ω + ω² = 0 ⇒ ω + ω² = −1 and ω³ = 1
= (x + y + z) ( x² + y² + z² − xy − yz − zx )
= x³ + y³ + z³ − 3xyz = L.H.S.
Using the formula:
(a + b + c)(a² + b² + c² − ab − bc − ca) = a³ + b³ + c³ − 3abc
Question 5 of Exercise 2.2 Class 10. Prove that
(1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸) … = 1 for 2n factors
Let, L.H.S. = (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸) … 2n factors
= (1 + ω)(1 + ω²)(1 + ω³)(1 + ω⁶) … 2n factors
∵ ω³ = 1 ⇒ ω⁶ = (ω³)² = 1
= (1 + ω)(1 + ω²)(1 + ω)(1 + ω²) … n factors
= [ (1 + ω)(1 + ω²) ]ⁿ
= [ 1 + ω² + ω + ω³ ]ⁿ
= [ 1 + ω + ω² + ω³ ]ⁿ
= [ 0 + 1 ]ⁿ
= 1ⁿ
= 1 = R.H.S. ∴ L.H.S. = R.H.S.
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 math to further enhance you concepts.