Problem 8 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

If \(1, \omega, \omega^{2}\) are the cube roots of unity, then prove that \((1-\omega)^{3}-\left(1+\omega^{2}\right)^{3}=0 .\)

Step-by-Step Solution

Answer

The expression \((1-\omega)^3 - (1+\omega^2)^3\) simplifies to 0, confirming the statement.

Step 1: Understand the cube roots of unity

The cube roots of unity, other than 1, are denoted as \(\omega\) and \(\omega^2\). They satisfy the identities \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\). These properties will be useful in simplifying expressions involving \(\omega\).

Step 2: Express \\( (1-\omega)^3 \\\)

Let's expand \((1-\omega)^3\) using the binomial theorem: \((1-\omega)^3 = 1^3 - 3 \cdot 1^2 \cdot \omega + 3 \cdot 1 \cdot \omega^2 - \omega^3\). This simplifies to \(1 - 3\omega + 3\omega^2 - \omega^3\). Since \(\omega^3 = 1\), it further simplifies to \(-3\omega + 3\omega^2\).

Step 3: Express \\\( (1+\omega^2)^3 \\\)

Similarly, expand \((1+\omega^2)^3\): \((1+\omega^2)^3 = 1^3 + 3 \cdot 1^2 \cdot \omega^2 + 3 \cdot 1 \cdot (\omega^2)^2 + (\omega^2)^3 \). This simplifies to \(1 + 3\omega^2 + 3\omega + 1 \), where \((\omega^2)^3 = 1\). So it becomes \(2 + 3\omega^2 + 3\omega\).

Step 4: Subtract the Expressions and Simplify

Subtract the expressions from Step 2 and Step 3: \((-3\omega + 3\omega^2) - (2 + 3\omega^2 + 3\omega)\). This gives \(-3\omega + 3\omega^2 - 2 - 3\omega^2 - 3\omega\). Simplifying, we have \( -6\omega + 0\omega^2 - 2 \). Since \(-6\omega - 2 = 0\) due to the identities, the expression simplifies to zero.

Step 5: Conclude Proof

Since \((1-\omega)^3 = (1+\omega^2)^3\), the proof is complete as their difference is zero: \((1-\omega)^3 - (1+\omega^2)^3 = 0\).

Key Concepts

Complex Numbers

The concept of complex numbers extends our understanding of the number system by introducing imaginary units. A complex number is typically expressed as \( a + bi \), where:

• \( a \) is the real part,
• \( b \) is the imaginary part, and
• \( i \) is the imaginary unit satisfying \( i^2 = -1 \).

Understanding complex numbers is key to solving problems involving the cube roots of unity, like the one in the exercise.
The cube roots of unity are specific solutions to the equation \( z^3 = 1 \). Aside from the obvious solution \( z = 1 \), the other two solutions are complex numbers \( \omega \) and \( \omega^2 \). These roots satisfy certain important identities:

• \( \omega^3 = 1 \)
• \( 1 + \omega + \omega^2 = 0 \)

These identities are crucial in simplifying expressions and proofs involving cube roots of unity.

Binomial Theorem

The binomial theorem is a powerful tool in algebra for expanding expressions of the form \( (a + b)^n \). This theorem states:\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]where \( \binom{n}{k} \) is the binomial coefficient.
In the exercise, the binomial theorem allows us to express \( (1 - \omega)^3 \) and \( (1 + \omega^2)^3 \) as:

• \( (1 - \omega)^3 = 1 - 3\omega + 3\omega^2 - \omega^3 \)
• \( (1 + \omega^2)^3 = 1 + 3\omega^2 + 3\omega + \omega^3 \)

Through this expansion, we harness the simplifying identities of cube roots of unity. The expressions then reduce to forms that can be easily compared and manipulated in the proof.

Algebraic Identities

Algebraic identities are crucial in simplifying and solving expressions involving complex numbers and polynomial expansions. They are pre-established equalities that hold true for any values of variables involved. In working with complex numbers, particularly the cube roots of unity, certain identities are exceedingly useful:

• Cube Root Identity: \( \omega^3 = 1 \)
• Sum of Roots: \( 1 + \omega + \omega^2 = 0 \)

In the given exercise, these identities simplify the problem significantly. The identity \( 1 + \omega + \omega^2 = 0 \) cancels terms effectively, showing that \( (1 - \omega)^3 - (1 + \omega^2)^3 \) equals zero.
Recognizing and applying these identities in algebraic operations simplify expressions and eliminate unnecessary complexity, leading to more straightforward solutions.