Q. \[ x^3 – 27 \]

Q: Factor x^3-27 completely?

Use difference of cubes: \,x^3-27=x^3-3^3=(x-3)(x^2+3x+9).

Q: What are the real solutions to x^3-27=0?

Solve x^3=27. Real solution: x=3. The other roots from x^2+3x+9=0 are complex.

Q: Find all complex roots of x^3-27=0.

Roots: x=3 and solutions of x^2+3x+9=0. Compute x=\frac{-3\pm\sqrt{9-36}}{2}=\frac{-3\pm3i\sqrt{3}}{2}.

Q: Graph/intercepts of y=x^3-27 ?

y-intercept: y(0)=-27. x-intercept: set x^3-27=0\Rightarrow x=3. Only one real crossing at (3,0).

Q: Solve x^3=27 without factoring.

Take cube root: x=\sqrt[3]{27}=3. Over reals, this is the only solution because cube root is single-valued.

Q: Determine whether x^3-27 is divisible by x-3.

Yes. Since 3^3-27=0, (x-3) is a factor by the Factor Theorem.

Answer

We factor the difference of cubes.

\[
x^3-27=x^3-3^3=(x-3)(x^2+3x+9)
\]

The real solution comes from \(x-3=0\).

\[
x=3
\]

Since \(x^2+3x+9\) has discriminant \(3^2-4\cdot1\cdot9= -27<0\), it has no real roots.

Final result: \(x^3-27=(x-3)(x^2+3x+9)\), and the only real solution is \(x=3\).

Detailed Explanation

We want to work with the expression \(x^3 – 27\). A key goal in many algebra problems is to factor an expression. This one is a difference of cubes.

Step 1: Recognize the pattern

Recall the difference of cubes identity:

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

Match it to \(x^3 – 27\). Notice that \(x^3\) is a cube already, and \(27\) is also a cube because \(27 = 3^3\).

So we can rewrite:

\[
x^3 – 27 = x^3 – 3^3.
\]

Step 2: Apply the identity

Let \(a = x\) and \(b = 3\). Substitute into the formula:

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

Step 3: Simplify inside the parentheses

Compute each term:

\(x^2\) stays as \(x^2\)
\(x\cdot 3 = 3x\)
\(3^2 = 9\)

So the expression becomes:

\[
(x – 3)(x^2 + 3x + 9).
\]

Step 4: Final answer

The factored form of \(x^3 – 27\) is:

\[
x^3 – 27 = (x – 3)(x^2 + 3x + 9).
\]

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Algebra FAQ

Factor \(x^3-27\) completely?

Use difference of cubes: \(\,x^3-27=x^3-3^3=(x-3)(x^2+3x+9)\).

What are the real solutions to \(x^3-27=0\)?

Solve \(x^3=27\). Real solution: \(x=3\). The other roots from \(x^2+3x+9=0\) are complex.

Find all complex roots of \(x^3-27=0\).

Roots: \(x=3\) and solutions of \(x^2+3x+9=0\). Compute \(x=\frac{-3\pm\sqrt{9-36}}{2}=\frac{-3\pm3i\sqrt{3}}{2}\).

Graph/intercepts of \(y=x^3-27\) ?

\(y\)-intercept: \(y(0)=-27\). \(x\)-intercept: set \(x^3-27=0\Rightarrow x=3\). Only one real crossing at \((3,0)\).

Solve \(x^3=27\) without factoring.

Take cube root: \(x=\sqrt[3]{27}=3\). Over reals, this is the only solution because cube root is single-valued.

Determine whether \(x^3-27\) is divisible by \(x-3\).

Yes. Since \(3^3-27=0\), \((x-3)\) is a factor by the Factor Theorem.

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