To follow up, if we subtract the former equation from the latter, we obtain:
[MATH]\left(\frac{p}{q}\right)^3-1=0[/MATH]
Thus:
[MATH]\left(\frac{p}{q}\right)^3=1[/MATH]
Another approach would be to begin at the same starting point:
[MATH](p+q)^2=pq[/MATH]
Expand and collect like terms:
[MATH]p^2+pq+q^2=0[/MATH]
We should recognize this could be a factor in a difference of cubes. Observing that in the original equation \(p\ne q\) we may multiply by \(p-q\) to get:
[MATH](p-q)(p^2+pq+q^2)=0[/MATH]
And so:
[MATH]p^3-q^3=0[/MATH]
Divide by \(q^3\):
[MATH]\left(\frac{p}{q}\right)^3-1=0[/MATH]
Thus:
[MATH]\left(\frac{p}{q}\right)^3=1[/MATH]