Q. \(x^3 – 8 = (x-2)(x^2 + 2x + 4)\)

We want to factor the expression \(x^3 – 8\).

Step 1: Recognize the type of expression.

Notice that \(x^3 – 8\) has the form of a difference of cubes:

\[
x^3 – 8 = a^3 – b^3
\]

Match terms:

• \(a^3 = x^3\), so \(a = x\)
• \(b^3 = 8\). Since \(2^3 = 8\), we have \(b = 2\)

Step 2: Use the difference of cubes formula.

The factoring formula is:

\[
a^3 – b^3 = (a – b)(a^2 + ab + b^2)
\]

Step 3: Substitute \(a = x\) and \(b = 2\).

\[
x^3 – 2^3 = (x – 2)(x^2 + x\cdot 2 + 2^2)
\]

Step 4: Simplify the trinomial inside the parentheses.

Compute each part:

• \(x\cdot 2 = 2x\)
• \(2^2 = 4\)

So:

\[
x^3 – 8 = (x – 2)(x^2 + 2x + 4)
\]

Step 5: Check if the quadratic can be factored further.

Consider \(x^2 + 2x + 4\). Its discriminant is:

\[
\Delta = 2^2 – 4\cdot 1 \cdot 4 = 4 – 16 = -12
\]

Because the discriminant is negative, it does not factor over the real numbers (and also not over the integers).

Final Answer:

\[
x^3 – 8 = (x – 2)(x^2 + 2x + 4)
\]