Q. \(x^3+64\)

Q: What are the factors of x^3+64?

Use 64=4^3: x^3+4^3=(x+4)(x^2-4x+16).

Q: Solve x^3+64=0.

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

Q: What are the complex roots of x^3+64=0?

From x^2-4x+16=0: x=\frac{4\pm\sqrt{-48}}{2}=2\pm 2\sqrt{3}\,i.

Q: How do you factor x^3+64 using the sum of cubes formula?

Apply a^3+b^3=(a+b)(a^2-ab+b^2) with a=x, b=4: x^3+64=(x+4)(x^2-4x+16).

Q: Is there a real root besides x=-4 for x^3+64=0?

No. The quadratic factor has discriminant b^2-4ac=16-64=-48, giving complex conjugate roots only.

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

f'(x)=3x^2. Setting f'(x)=0 gives x=0 (double root), the only critical point.

Answer

We want to factor \(x^3+64\). Note that \(64=4^3\), so \(x^3+4^3\) is a sum of cubes:

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

Final result: \((x+4)(x^2-4x+16)\).

Detailed Explanation

We are given the expression \(x^3+64\). This is a sum of two cubes, because \(64\) can be written as \(4^3\):

\[64 = 4^3\]

So the expression becomes:

\[x^3+64 = x^3+4^3\]

Now we use the sum of cubes factorization formula:

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

Here, we identify \(a=x\) and \(b=4\). Substitute into the formula:

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

Now simplify inside the parentheses step by step. First compute \(x\cdot 4=4x\), and compute \(4^2=16\):

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

Therefore, the factorization of \(x^3+64\) is:

\[\boxed{(x+4)(x^2-4x+16)}\]

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Algebra FAQ

What are the factors of \(x^3+64\)?

Use \(64=4^3\): \(x^3+4^3=(x+4)(x^2-4x+16)\).

Solve \(x^3+64=0\).

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

What are the complex roots of \(x^3+64=0\)?

From \(x^2-4x+16=0\): \(x=\frac{4\pm\sqrt{-48}}{2}=2\pm 2\sqrt{3}\,i\).

How do you factor \(x^3+64\) using the sum of cubes formula?

Apply \(a^3+b^3=(a+b)(a^2-ab+b^2)\) with \(a=x\), \(b=4\): \(x^3+64=(x+4)(x^2-4x+16)\).

Is there a real root besides \(x=-4\) for \(x^3+64=0\)?

No. The quadratic factor has discriminant \(b^2-4ac=16-64=-48\), giving complex conjugate roots only.

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

\(f'(x)=3x^2\). Setting \(f'(x)=0\) gives \(x=0\) (double root), the only critical point.

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