Q. \(\,x^3-64\,\)

Q: How do you factor x^3-64?

\[x^3-64=x^3-4^3=(x-4)(x^2+4x+16).\]

Q: What are the real roots of x^3-64=0?

\[x^3=64 \Rightarrow x=4.\] The other factor x^2+4x+16 has discriminant 16-64=-48<0, so no more real roots.

Q: How do you solve x^3-64=0 over complex numbers?

\[x-4=0 \Rightarrow x=4.\] And x^2+4x+16=0 gives x=\frac{-4\pm\sqrt{-48}}{2}=-2\pm 2\sqrt{3}\,i.

Q: How can you use the sum/difference of cubes formula here?

Use a^3-b^3=(a-b)(a^2+ab+b^2). Let a=x, b=4: \[x^3-4^3=(x-4)(x^2+4x+16).\]

Q: Why does x^2+4x+16 not factor over the reals?

Its discriminant is \Delta=4^2-4\cdot1\cdot16=16-64=-48<0, so it has no real roots and cannot factor into real linear terms.

Q: What is the derivative of f(x)=x^3-64 and where is it critical?

f'(x)=3x^2. The critical point satisfies 3x^2=0, so x=0. Then f(0)=-64.

Answer

We factor the difference of cubes.

\[
x^3 – 64 = x^3 – 4^3 = (x-4)(x^2+4x+16)
\]

Since \(x^2+4x+16\) has discriminant \(16-64=-48<0\), the only real root comes from \(x-4=0\).

\[
x=4
\]

Final result:

\[
x^3-64=(x-4)(x^2+4x+16), \quad \text{real solution } x=4
\]

Detailed Explanation

We want to factor the expression \(x^3-64\).

Step 1: Recognize a special pattern

The expression \(x^3-64\) matches the difference of cubes pattern:

\[
a^3-b^3 = (a-b)\left(a^2+ab+b^2\right)
\]

Here, \(x^3\) is a cube, and \(64\) is also a cube because \(64=4^3\).

So we identify \(a=x\) and \(b=4\).

Step 2: Apply the difference of cubes formula

Substitute \(a=x\) and \(b=4\) into the formula:

\[
x^3-4^3 = (x-4)\left(x^2+x\cdot 4+4^2\right)
\]

Now simplify each part.

Step 3: Simplify inside the parentheses

Compute the terms in \(\left(x^2+x\cdot 4+4^2\right)\):

\[
x\cdot 4 = 4x,\quad 4^2 = 16
\]

So the expression becomes:

\[
x^3-64 = (x-4)\left(x^2+4x+16\right)
\]

Final Answer

\[
x^3-64 = (x-4)\left(x^2+4x+16\right)
\]

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Algebra FAQ

How do you factor \(x^3-64\)?

\[x^3-64=x^3-4^3=(x-4)(x^2+4x+16).\]

What are the real roots of \(x^3-64=0\)?

\[x^3=64 \Rightarrow x=4.\] The other factor \(x^2+4x+16\) has discriminant \(16-64=-48<0\), so no more real roots.

How do you solve \(x^3-64=0\) over complex numbers?

\[x-4=0 \Rightarrow x=4.\] And \(x^2+4x+16=0\) gives \(x=\frac{-4\pm\sqrt{-48}}{2}=-2\pm 2\sqrt{3}\,i.\)

How can you use the sum/difference of cubes formula here?

Use \(a^3-b^3=(a-b)(a^2+ab+b^2).\) Let \(a=x\), \(b=4\): \[x^3-4^3=(x-4)(x^2+4x+16).\]

Why does \(x^2+4x+16\) not factor over the reals?

Its discriminant is \(\Delta=4^2-4\cdot1\cdot16=16-64=-48<0,\) so it has no real roots and cannot factor into real linear terms.

What is the derivative of \(f(x)=x^3-64\) and where is it critical?

\(f'(x)=3x^2\). The critical point satisfies \(3x^2=0\), so \(x=0\). Then \(f(0)=-64\).

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