+130466 Sasini did the binomial expansion of (p - 3)^3
I think what you might have asked for is to either factor, or find the roots of, p^3 - 27...I'll do both....
Factoring is just the difference of two cubes...so we have
(p - 3) * (p^2 + 3p + 9)
We can set these factors equal to 0 to find the roots......setting (p - 3) = 0 gives us that p = 3.
The second factor - (p^2 + 3p + 9) - set to 0, has no real solutions (roots).
I think that's it......
+752 p^3-(3^3)
=p^3-(3*P^2*3)+(3*p*3^2)-3^3
=p^3-9p^2+27p-27
I think u can uderstand about it. 
+130466 Sasini did the binomial expansion of (p - 3)^3
I think what you might have asked for is to either factor, or find the roots of, p^3 - 27...I'll do both....
Factoring is just the difference of two cubes...so we have
(p - 3) * (p^2 + 3p + 9)
We can set these factors equal to 0 to find the roots......setting (p - 3) = 0 gives us that p = 3.
The second factor - (p^2 + 3p + 9) - set to 0, has no real solutions (roots).
I think that's it......
