Break up 8 into two summands such that the sum of their cubes is the least possible.

To break up \( 8 \) into two summands, say \( x \) and \( y \), we want to minimize the sum of their cubes. The problem can be expressed as:

\[
x + y = 8
\]

And we want to minimize:

\[
S(x, y) = x^3 + y^3
\]

By substituting \( y = 8 - x \) into the equation, we get:

\[
S(x) = x^3 + (8 - x)^3
\]

Now, we differentiate \( S(x) \) to find the value of \( x \) that minimizes the sum:

\[
\frac{dS(x)}{dx} = 3x^2 - 3(8 - x)^2
\]

Set the derivative to zero to find the critical points:

\[
3x^2 = 3(8 - x)^2
\]

Dividing by 3:

\[
x^2 = (8 - x)^2
\]

Take the square root of both sides:

\[
x = 8 - x
\]

This simplifies to:

\[
2x = 8 \quad \Rightarrow \quad x = 4
\]

Thus, the two summands are \( x = 4 \) and \( y = 8 - x = 4 \).

Finally, we calculate the sum of their cubes:

\[
S(4, 4) = 4^3 + 4^3 = 64 + 64 = 128
\]

Therefore, the least possible sum of cubes when breaking up 8 into two summands is \( \boxed{128} \), with both summands being \( 4 \).

Would you like further details or explanations?

Here are 5 related questions:
1. What if we split \( 8 \) into summands of unequal values, such as \( x = 5 \) and \( y = 3 \)?
2. How would this problem change if you had to minimize the sum of their squares instead of cubes?
3. Can this approach be generalized to breaking up any number \( n \) into two summands to minimize their cube sum?
4. How does this minimization relate to calculus concepts like critical points and derivatives?
5. What would happen if you had to break up \( 8 \) into more than two summands?

**Tip:** When solving optimization problems, always check if the result satisfies the conditions by verifying critical points using calculus!

Explore how to split the number 8 into two summands to minimize the sum of their cubes. Using algebra and calculus, we find that the least possible sum is 128 with both summands being 4. Suitable for students in Grades 10-12.

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