If $p^{2}+q^{2}-29=2 p q-20=52-2 p q$, then the difference between the maximum and minimum possible value of $\left(p^{3}-q^{3}\right)$ is

We can see we have a chain of equal expressions that might seem confusing at first glance. Let us break this down so you can easily follow the logic.

We're told that: $p^{2}+q^{2}-29=2 p q-20=52-2 p q$

This means all three expressions are equal to each other. Let's use this fact strategically.

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Since $2pq - 20 = 52 - 2pq$, let's solve for $pq$:

$2pq - 20 = 52 - 2pq$

$4pq - 20 = 52$

$4pq = 72$

Therefore: $pq = 18$

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Now we use the fact that $p^{2}+q^{2}-29=2 p q-20$.

Since we know $pq = 18$, we can substitute $2pq = 36$:

$p^{2}+q^{2}-29 = 36-20 = 16$

Therefore: $p^{2}+q^{2} = 16 + 29 = 45$

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Here's where we use a crucial algebraic identity:

$p^{3}-q^{3}=(p-q)(p^{2}+pq+q^{2})$

Why this identity works: Think of it as factoring. Just like $a^2 - b^2 = (a-b)(a+b)$, we have a similar pattern for cubes.

Substituting our known values:

$p^{3}-q^{3}=(p-q)(45+18)=63(p-q)$

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We need to find $(p-q)^2$ using another key identity:

$(p-q)^{2} = p^{2} + q^{2} - 2pq$

Why this works: $(p-q)^{2} = (p-q)(p-q) = p^{2} - 2pq + q^{2}$

Substituting our values:

$(p-q)^{2} = 45 - 2(18) = 45 - 36 = 9$

Therefore: $p-q = \pm 3$

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Since $p^{3}-q^{3} = 63(p-q)$:

Maximum value: When $p-q = 3$, we get $p^{3}-q^{3} = 63 \times 3 = 189$

Minimum value: When $p-q = -3$, we get $p^{3}-q^{3} = 63 \times (-3) = -189$

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Difference = Maximum value - Minimum value = $189 - (-189) = 189 + 189 = 378$

Key takeaways for future problems:

When you have multiple equal expressions, use them strategically to find individual values

Remember the identity: $p^{3}-q^{3}=(p-q)(p^{2}+pq+q^{2})$

Remember the identity: $(p-q)^{2} = p^{2} + q^{2} - 2pq$

When finding maximum and minimum values, consider both positive and negative cases

Answer: 378