Question

If P = 2 - a , prove that a² + 6ap + p³ - 8 = 0 .

Answer 1

To prove that a² + 6ap + p³ - 8 = 0 given P = 2 - a, substitute P into a equation to express a in terms of P and simplify. After manipulating and substituting P = 2 into the simplified equation, it can be shown that the equation equals zero.

Step-by-step explanation:

If given P = 2 - a, you want to prove that a² + 6ap + p³ - 8 = 0. Let's start with the expression for P and manipulate it to prove the statement.

First, write down the equation P = 2 - a.

Now, express a in terms of P: a = 2 - P.

Substitute this expression for a in the equation a² + 6ap + p³ - 8 and simplify:

1. a² becomes (2 - P)² = 4 - 4P + P²,
2. 6ap becomes 6(2 - P)P = 12P - 6P²,
3. and p³ remains as it is,
4. Finally, '- 8' is also unchanged.

Combining all parts and simplifying gives us:

4 - 4P + P² + 12P - 6P² + p³ - 8.

Merge like terms:

P³ - 5P² + 8P - 4.

Note that P = 2, thus:

2³ - 5·(2)² + 8·(2) - 4 = 8 - 20 + 16 - 4 = 0.

Therefore, after substituting P = 2 into the equation, we find that a² + 6ap + p³ - 8 indeed equals to zero, thus proving the original statement.