Question

If \(a+b+c=3, a^{2}+b^{2}+c^{2}=6\) and \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\) where \(a, b, c\) are all non-zero, then \(a b c\) is : (a) \(\frac{1}{3}\) (b) \(\frac{2}{3}\) (c) \(\frac{3}{2}\) (d) 1

Short answer

1) \(a+b+c=3\) 2) \(a^2+b^2+c^2=6\) 3) \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\) Answer: (c) \(\frac{3}{2}\)

Step-by-step solution

Observe the given equations

The given equations are: 1) \(a+b+c=3\) 2) \(a^2+b^2+c^2=6\) 3) \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\) We need to find the value of \(abc\).

Simplify the third equation

To make the third equation more manageable, we can find a common denominator, which in this case would be \(abc\). So, the equation becomes: \(\frac{bc+ac+ab}{abc}=1\) This can be further simplified to: \(bc+ac+ab=abc\)

Square the first equation

To find a relationship between the first and second equation, we can square the first equation: \((a+b+c)^2 = 3^2\) Expanding this gives us: \(a^2+b^2+c^2+2(ab+ac+bc) = 9\)

Substitute the second equation

Now, we can substitute the second equation (\(a^2+b^2+c^2=6\)) into the expanded version of the first equation: \(6 + 2(ab+ac+bc) = 9\) Solving for \((ab+ac+bc)\), we get: \(ab+ac+bc = \frac{3}{2}\)

Substitute the expression for \(abc\)

Now, we can substitute the expression for \(abc\) that we found in Step 2: \(bc+ac+ab=abc \Rightarrow \frac{3}{2} = abc\)

Identify the correct answer

From the final equation \(\frac{3}{2}=abc\), we can see that the correct answer is: (c) \(\frac{3}{2}\)

Key concepts

Quadratic Equations

At the heart of algebra, quadratic equations represent a fundamental concept that students often tackle. A quadratic equation is an algebraic expression of the second degree, typically in the form of \( ax^2 + bx + c = 0 \), where \( a, b, \text{ and } c \) are constants, and \( a \) is not equal to zero. The solutions to these equations, also known as roots, can be found using various methods such as factoring, completing the square, or applying the quadratic formula

Relevance to the Exercise

In the given exercise, although we are not dealing with a standard quadratic equation, the process of squaring the first equation to relate it to the second is a technique often employed in solving quadratics. By manipulating the equations, we strive to express variables in a form where their relationships become clear. This is akin to how one might manipulate a quadratic equation to make it factorable or suitable for application of the quadratic formula.

System of Equations

A system of equations consists of two or more equations set equal to one another with the same variables. The goal is to find the values of the variables that satisfy all equations simultaneously. These systems can be linear, quadratic, or even exponential, and are solved using methods like substitution, elimination, or graphically.

Techniques in Practice

In our exercise, we are looking at a system of three algebraic equations. The steps involve manipulating these equations to isolate \( abc \) as the sole variable. This is done through substitution where one equation is solved in terms of others. Effective manipulation of these systems requires a firm understanding of algebraic operations and the relationships between equations within a system.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations. For instance, \( a + 7 \) and \( bc \) are simple expressions, while \( a^2 + b^2 + c^2 \) is more complex. Expressions become equations when paired with an equal sign and a value.

Expression Manipulation

As demonstrated in the exercise's Step 2, simplifying an algebraic expression is crucial. Simplification can involve finding a common denominator or factoring. The objective is to rewrite the expression so that it unveils a relationship or value useful in solving a problem, as we saw in the transformation of \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 \) to \( abc \) to find the final solution.