Finding the Value of \(a^3 + b^3 + c^3\) Using Algebraic Identity

The problem asks us to find the value of the expression \(a^3 + b^3 + c^3\) given the values of a, b, and c.

We are given:

• \(a = 12\)
• \(b = -8\)
• \(c = -4\)

We need to calculate \(a^3 + b^3 + c^3\).

Applying an Important Algebraic Identity

There is a useful algebraic identity related to the sum of cubes:

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

A special case of this identity occurs when \(a+b+c = 0\). In this case, the right side of the identity becomes \(0 \times (a^2 + b^2 + c^2 - ab - bc - ca) = 0\).

So, if \(a+b+c = 0\), the identity simplifies to:

\(a^3 + b^3 + c^3 - 3abc = 0\)

Which can be rearranged as:

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

Checking the Condition \(a+b+c = 0\)

Let's check if the sum of the given values of a, b, and c is equal to 0:

\(a + b + c = 12 + (-8) + (-4)\)

\(a + b + c = 12 - 8 - 4\)

\(a + b + c = 12 - (8 + 4)\)

\(a + b + c = 12 - 12\)

\(a + b + c = 0\)

Since \(a+b+c = 0\), we can use the simplified identity \(a^3 + b^3 + c^3 = 3abc\).

Calculating \(3abc\)

Now, we substitute the given values of a, b, and c into the expression \(3abc\):

\(3abc = 3 \times (12) \times (-8) \times (-4)\)

\(3abc = 36 \times (-8) \times (-4)\)

Multiplying the negative numbers:

\((-8) \times (-4) = 32\)

So, the expression becomes:

\(3abc = 36 \times 32\)

Let's perform the multiplication:

Thus,

\(36 \times 32 = 1152\)

Final Value of \(a^3 + b^3 + c^3\)

Since \(a+b+c = 0\), we found that \(a^3 + b^3 + c^3 = 3abc\).

We calculated \(3abc = 1152\).

Therefore, \(a^3 + b^3 + c^3 = 1152\).

Revision Table - Sum of Cubes Calculation

Additional Information - Algebraic Identities

Algebraic identities are equations that are true for all possible values of the variables involved. They are very useful for simplifying expressions and solving equations.

Some other common algebraic identities include:

• \((x+y)^2 = x^2 + 2xy + y^2\)
• \((x-y)^2 = x^2 - 2xy + y^2\)
• \(x^2 - y^2 = (x+y)(x-y)\)
• \((x+y)^3 = x^3 + y^3 + 3xy(x+y)\)
• \((x-y)^3 = x^3 - y^3 - 3xy(x-y)\)
• \(x^3 + y^3 = (x+y)(x^2 - xy + y^2)\)
• \(x^3 - y^3 = (x-y)(x^2 + xy + y^2)\)

The identity \(a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)\) is particularly useful when dealing with sums of cubes of three variables. Recognizing the special case where the sum of the variables is zero can significantly simplify calculations, as seen in this problem.