Understanding the Algebraic Simplification Problem

The question asks us to evaluate a complex fraction involving variables \(a\), \(b\), \(c\), and newly defined terms \(\alpha\), \(\beta\), \(\gamma\). We are given the definitions of \(\alpha\), \(\beta\), and \(\gamma\) in terms of \(a\), \(b\), and \(c\).

The given definitions are:

• \(\alpha = a^2 - bc\)
• \(\beta = b^2 - ac\)
• \(\gamma = c^2 - ab\)

We need to find the value of the expression:

\(\rm \frac{a \alpha+b \beta+c \gamma}{(a+b+c)(\alpha+\beta+\gamma)}\)

Step-by-Step Calculation of the Numerator

Let's first calculate the numerator, which is \(a \alpha + b \beta + c \gamma\). We will substitute the given expressions for \(\alpha\), \(\beta\), and \(\gamma\):

\(a \alpha + b \beta + c \gamma = a(a^2 - bc) + b(b^2 - ac) + c(c^2 - ab)\)

Now, distribute \(a\), \(b\), and \(c\) into the parentheses:

\(= a \cdot a^2 - a \cdot bc + b \cdot b^2 - b \cdot ac + c \cdot c^2 - c \cdot ab\)

\(= a^3 - abc + b^3 - abc + c^3 - abc\)

Combine the terms:

\(= a^3 + b^3 + c^3 - 3abc\)

So, the numerator simplifies to \(a^3 + b^3 + c^3 - 3abc\).

Step-by-Step Calculation of the Denominator

The denominator is \((a+b+c)(\alpha+\beta+\gamma)\). Let's first calculate the sum \(\alpha+\beta+\gamma\):

\(\alpha + \beta + \gamma = (a^2 - bc) + (b^2 - ac) + (c^2 - ab)\)

\(= a^2 - bc + b^2 - ac + c^2 - ab\)

Rearrange the terms:

\(= a^2 + b^2 + c^2 - ab - bc - ac\)

Now, we need to calculate the full denominator:

\((a+b+c)(\alpha+\beta+\gamma) = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ac)\)

This expression matches a well-known algebraic identity:

\((x+y+z)(x^2+y^2+z^2-xy-yz-zx) = x^3+y^3+z^3-3xyz\)

Using this identity with \(x=a\), \(y=b\), and \(z=c\), we get:

\((a+b+c)(a^2 + b^2 + c^2 - ab - bc - ac) = a^3 + b^3 + c^3 - 3abc\)

So, the denominator also simplifies to \(a^3 + b^3 + c^3 - 3abc\).

Evaluating the Final Expression

Now we have the simplified forms for both the numerator and the denominator:

• Numerator: \(a^3 + b^3 + c^3 - 3abc\)
• Denominator: \(a^3 + b^3 + c^3 - 3abc\)

The fraction is:

\(\rm \frac{a \alpha+b \beta+c \gamma}{(a+b+c)(\alpha+\beta+\gamma)} = \frac{a^3 + b^3 + c^3 - 3abc}{a^3 + b^3 + c^3 - 3abc}\)

Assuming that the denominator \(a^3 + b^3 + c^3 - 3abc\) is not equal to zero, the fraction simplifies to 1.

The final value of the expression is 1.

The value of the expression \(\rm \frac{a \alpha+b \beta+c \gamma}{(a+b+c)(\alpha+\beta+\gamma)}\) is 1, provided the denominator is non-zero.

Revision Table: Key Algebraic Concepts

Additional Information: The \(a^3+b^3+c^3-3abc\) Identity

The identity \(a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ac)\) is a fundamental result in algebra. It has several important applications.

One significant application is related to the condition \(a^3+b^3+c^3 = 3abc\). From the identity, this condition is true if and only if:

\((a+b+c)(a^2+b^2+c^2-ab-bc-ac) = 0\)

This means either \(a+b+c = 0\) or \(a^2+b^2+c^2-ab-bc-ac = 0\).

The second condition \(a^2+b^2+c^2-ab-bc-ac = 0\) can be rewritten as:

\(\frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2) = 0\)

Since squares of real numbers are non-negative, the sum of squares is zero if and only if each term is zero, i.e., \(a-b=0\), \(b-c=0\), and \(c-a=0\). This implies \(a=b=c\).

Therefore, \(a^3+b^3+c^3 = 3abc\) if and only if \(a+b+c = 0\) or \(a=b=c\).

This identity was key to simplifying the denominator and showing it was equal to the numerator.