Bài 6:
a) \(P=27-27a+9x^2-x^3\)
\(P=3^3-3\cdot3^2\cdot x+3\cdot3\cdot x^2-x^3\)
\(P=\left(3-x\right)^3\)
Thay \(x=-17\) vào P ta được:
\(\left(3--17\right)^3=20^3=8000\)
Vậy: ...
b) \(Q=x^3+3x^2+3x\)
\(Q=x^3+3x^2+3x+1-1\)
\(Q=\left(x+3\right)^3-1\)
Thay \(x=99\) vào Q ta có:
\(\left(99+1\right)^3-1=100^3-1=1000000-1=999999\)
Vậy: ...
6:
a: P=(3-x)^3=(3+17)^3=20^3=8000
b: Q=x^3+3x^2+3x+1-1
=(x+1)^3-1
=(99+1)^3-1
=100^3-1
=999999