184 B
184 B
m^3=n^3+n^2p
let m=n+k , then
3n^2k+3nk^2+k^3=n^2p\\
(p-3k)n^2-3k^2n-k^3=0
witch p>3k . solving this function, we get
n = \frac{3k^2 \pm k\sqrt{4kp-3k^2}}{p-3k}
m^3=n^3+n^2p
let m=n+k , then
3n^2k+3nk^2+k^3=n^2p\\
(p-3k)n^2-3k^2n-k^3=0
witch p>3k . solving this function, we get
n = \frac{3k^2 \pm k\sqrt{4kp-3k^2}}{p-3k}