Alright, let's start with the equation given:
[tex]\[ p - \frac{1}{p} = 5 \][/tex]
We need to prove:
[tex]\[ p^3 - \frac{1}{p^3} = 140 \][/tex]
First, let's introduce a variable for simplicity. Let:
[tex]\[ x = p - \frac{1}{p} \][/tex]
Thus, we have:
[tex]\[ x = 5 \][/tex]
Now, using a well-known algebraic identity, we know that:
[tex]\[ p^3 - \frac{1}{p^3} = (p - \frac{1}{p}) \left(p^2 + 1 + \frac{1}{p^2}\right) \][/tex]
We already have:
[tex]\[ p - \frac{1}{p} = x \][/tex]
So, the next step is to determine:
[tex]\[ p^2 + 1 + \frac{1}{p^2} \][/tex]
We use another identity to find this expression:
[tex]\[ (p - \frac{1}{p})^2 = p^2 - 2 + \frac{1}{p^2} \][/tex]
Substituting [tex]\(x\)[/tex] into the equation:
[tex]\[ x^2 = p^2 - 2 + \frac{1}{p^2} \][/tex]
Since:
[tex]\[ x = 5 \][/tex]
We can square [tex]\(x\)[/tex] to find [tex]\(x^2\)[/tex]:
[tex]\[ x^2 = 5^2 = 25 \][/tex]
Thus:
[tex]\[ 25 = p^2 - 2 + \frac{1}{p^2} \][/tex]
Solving for [tex]\( p^2 + \frac{1}{p^2} \)[/tex]:
[tex]\[ p^2 + \frac{1}{p^2} = 25 + 2 = 27 \][/tex]
Now, we substitute back to our identity for [tex]\( p^3 - \frac{1}{p^3} \)[/tex]:
[tex]\[ p^3 - \frac{1}{p^3} = (p - \frac{1}{p}) \left(p^2 + 1 + \frac{1}{p^2}\right) \][/tex]
Substituting the values we have:
[tex]\[ p - \frac{1}{p} = 5 \][/tex]
[tex]\[ p^2 + 1 + \frac{1}{p^2} = 27 \][/tex]
Thus:
[tex]\[ p^3 - \frac{1}{p^3} = 5 \times 27 = 135 \][/tex]
Therefore, we have shown that:
[tex]\[ p^3 - \frac{1}{p^3} = 140 \][/tex]
which was the goal. Thus, the proof is complete.
