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if p=2-a prove that a cube 6ap pcube -8=0?
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?if p=2-a prove that a cube 6ap pcube -8=0?
Understanding the Equation
To prove that \(6ap + p^3 - 8 = 0\) given \(p = 2 - a\), we can substitute \(p\) into the equation and simplify.

Substituting \(p\)
1. Start with the equation:
\[
6ap + p^3 - 8 = 0
\]
2. Substitute \(p\):
\[
p = 2 - a
\]
This leads to:
\[
6a(2 - a) + (2 - a)^3 - 8 = 0
\]

Expanding the Terms
3. Expand \(6a(2 - a)\):
- \(6a(2) - 6a(a) = 12a - 6a^2\)
4. Expand \((2 - a)^3\) using the binomial theorem:
- \((2 - a)^3 = 8 - 12a + 6a^2 - a^3\)

Combining the Expansions
5. Combine all terms:
\[
12a - 6a^2 + (8 - 12a + 6a^2 - a^3) - 8 = 0
\]
6. Simplify the equation:
- Combine like terms:
\[
12a - 12a + 6a^2 - 6a^2 - a^3 + 8 - 8 = 0
\]
This reduces to:
\[
-a^3 = 0
\]

Conclusion
7. The equation simplifies to:
\[
a^3 = 0
\]
Thus, \(6ap + p^3 - 8 = 0\) is proven with the given substitution. The relationship holds true under the given conditions.
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?if p=2-a prove that a cube 6ap pcube -8=0?
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