Problem 200 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the cubes roots of unity are \(1, \omega, \omega^{2}\) then the roots of the equation \((x-1)^{3}+8=0\), are (A) \(-1,-1+2 \omega,-1-\omega^{2}\) (B) \(-1,-1,-1\) (C) \(-1,1-2 \omega, 1-2 \omega^{2}\) (D) \(-1,1+2 \omega, 1+2 \omega^{2}\)

Step-by-Step Solution

Answer

The correct answer is (C).

Step 1: Simplify the Equation

Start with the equation \((x-1)^{3}+8=0\). To simplify, move 8 to the other side: \((x-1)^3 = -8\).

Step 2: Take the Cube Root

To solve \((x-1)^3 = -8\), take the cube root of both sides: \(x-1 = \sqrt[3]{-8}\). The cube root of \(-8\) is \(-2\). So, \(x-1 = -2\).

Step 3: Solve for x

From \(x-1 = -2\), add 1 to both sides to solve for \(x\): \(x = -1\).

Step 4: Consider Complex Roots

For the cube root of \(-8\), consider multiplying \(-2\) by cube roots of unity which are \(1, \omega, \omega^2\). This gives potential roots: \(-2 \times 1, -2 \times \omega, -2 \times \omega^2\).

Step 5: Add Back 1 for Each Root

Since the substitution was \(x-1\), add 1 back to each potential root to find the actual roots of the original equation: - For \(-2 \times 1\), add 1: \(-2 + 1 = -1\).- For \(-2 \times \omega\), add 1: \(-2\omega + 1 = 1 - 2\omega\).- For \(-2 \times \omega^2\), add 1: \(-2\omega^2 + 1 = 1 - 2\omega^2\).

Step 6: Identify the Correct Answer

Compare the roots \(-1, 1 - 2\omega, 1 - 2\omega^2\) with the provided options: (A) \(-1,-1+2\omega,-1-\omega^{2}\)(B) \(-1,-1,-1\)(C) \(-1,1-2\omega, 1-2\omega^{2}\)(D) \(-1,1+2\omega, 1+2\omega^{2}\)Option (C) matches the roots \(-1, 1 - 2\omega, 1 - 2\omega^2\).

Key Concepts

Cube Roots of Unity

Cube roots of unity are special numbers in complex mathematics. They are roots of the equation \(x^3 = 1\). When calculated, these roots are \(1, \omega, \omega^2\), where \(\omega\) is a complex number known as a primitive cube root of unity. It's important to remember:

• \(1\) is the real cube root of unity.
• \(\omega\) represents a complex root and is equal to \(\frac{-1 + \sqrt{3}i}{2}\).
• \(\omega^2\) is also a complex root and equals \(\frac{-1 - \sqrt{3}i}{2}\).

These complex numbers are special because they satisfy the equations \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\). This property is useful when dealing with polynomial roots involving unity powers.

Polynomial Equations

Polynomial equations are mathematical expressions that can consist of variables, coefficients, and exponents. A general form is \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0\).In the context of cube roots of unity, solving polynomial equations involves finding the roots that satisfy the equation. In our exercise, we started with the equation \((x-1)^3 + 8 = 0\), and simplified it to \((x-1)^3 = -8\). By taking the cube root on both sides, we obtain roots which include complex numbers multiplied by cube roots of unity.Understanding polynomial equations is essential because they help to:

• Identify all possible roots (real and complex).
• Utilize symmetry and properties of complex roots efficiently.
• Simplify solving processes using identities like cube roots of unity.

Complex Roots

Understanding complex roots is crucial when solving polynomial equations that yield non-real solutions. In the context of our exercise, after simplifying \((x-1)^3 = -8\), we get \(x - 1 = -2\). Multiplying \(-2\) by the cube roots of unity \(1, \omega, \omega^2\), the potential roots become:

• \(-2 \times 1 = -2\)
• \(-2 \times \omega = -2\omega\)
• \(-2 \times \omega^2 = -2\omega^2\)

Adding 1 back to each, we find the roots for the original equation:

• For \(-2\), the root is \(-1\).
• For \(-2\omega\), the root becomes \(1 - 2\omega\).
• For \(-2\omega^2\), the root transforms to \(1 - 2\omega^2\).

Complex roots are not as intuitive as real roots but are equally important. They allow us to solve equations that don't have real number solutions, opening up a world of possibilities in calculations and real-world applications.