Question

Show that \((x-5)^{3}\) is congruent to \(\left(x^{3}-5\right)\) modulo 3.

Short answer

By expanding and simplifying the left side of the equation and applying modulo operation, we derived that \(x^3 - 2\) is congruent to \(x^{3}-5\) modulo 3, which shows that \((x-5)^{3}\) is indeed congruent to \(\left(x^{3}-5\right)\) modulo 3.

Step-by-step solution

Expand the left side

Start by expanding the left-side expression \((x-5)^3\). The binomial formula yields: \(x^3 - 15x^2 + 75x - 125\).

Apply the modulus operation

Next, apply modulo 3 to each term separately: \(x^3 \mod 3 - 15x^2 \mod 3 + 75x \mod 3 - 125 \mod 3\). The remainder when \(x^3\), \(15x^2\), and \(75x\) are divided by 3 are \(x^3\), \(0\), and \(0\), respectively. The remainder when 125 is divided by 3 is 2, so we are then left with \(x^3 - 2\).

Simplify the expression

After simplifying, we are left with \(x^3 - 2\). However, because we are working with modulo 3, \(-2 \mod 3\) is equivalent to \(1 \mod 3\). Thus, our simplified expression matches the right side of the original problem: \(x^3 - 5 \mod 3 = x^3 - 2 = x^3 + 1\) as needed.

Key concepts

Congruence Relation

Congruence relation is a fundamental concept in modular arithmetic. It describes a scenario where two numbers have the same remainder when divided by a certain number, called the modulus. When we say that two numbers, say \( a \) and \( b \), are congruent modulo \( n \), we write it as \( a \equiv b \pmod{n} \). This means that the difference \( a-b \) is a multiple of \( n \). For example, if \( 7 \equiv 1 \pmod{3} \), it means that when you divide 7 by 3, and 1 by 3, both leave the same remainder of 1. Congruence relations are useful in simplifying calculations, especially when dealing with large numbers or complex expressions, as in polynomial equations. In the original exercise, when we say \( (x-5)^3 \equiv x^3 - 5 \pmod{3} \), we are stating that both expressions leave the same remainder when divided by 3.

Binomial Expansion

The binomial expansion allows us to expand expressions raised to a power. It uses the Binomial Theorem, which states that \((a + b)^n\) can be expanded into a sum involving terms of the form \( C(n, k) \cdot a^{n-k} \cdot b^k \), where \( C(n, k) \) is a binomial coefficient. In our problem, the expression \((x-5)^3\) is expanded as follows:

• First term: \(x^3\)
• Second term: \(-3 \cdot x^2 \cdot 5\)
• Third term: \(3 \cdot x \cdot 5^2\)
• Fourth term: \(-5^3\)

The expanded form \(x^3 - 15x^2 + 75x - 125\) uses the pattern: \((x^3) - 3(x^2)(5) + 3(x)(25) - 125\). This expansion is crucial for performing the modulo operation in the exercise.

Modulo Operation

The modulo operation is a way of finding the remainder after division of one number by another. In simpler terms, it gives you what's left when you break a number down into equal parts. Mathematically, \( a \mod n \) yields the remainder of \( a \) divided by \( n \). If \( a = qn + r \), with \( r \) being the remainder, then \( r \equiv a \pmod{n} \). In the solution, each term of the expanded polynomial is taken \( \mod 3 \):

• \(x^3 \mod 3\)
• \(-15x^2 \mod 3 = 0\)
• \(75x \mod 3 = 0\)
• \(-125 \mod 3 = 2 \)

Thus, the simplified form after applying modulo operation is \(x^3 - 2\), which is then adjusted for equivalency to \(x^3 + 1\) \(\pmod{3}\). This shows how each term affects the final remainder.

Polynomial Simplification

Polynomial simplification involves combining like terms and reducing expressions to their simplest forms. In conjunction with the modulo operation, some terms may drop out, or convert to simpler forms, especially when they result in full multiples of the modulus.For instance, in the exercise, terms like \(-15x^2\) and \(75x\) simplify to 0 after applying modulo 3.Additionally, think of simplifying as cleaning up an equation: unnecessary elements are removed, while the core structure remains. In this problem, understanding the relationship between different terms and the modulus allows us to simplify the original polynomial \((x-5)^3\) to a form congruent with \(x^3 - 5\), making calculations much easier.