a, \(x^3+3x^2+3x+6\)
\(=x^3+3x^2\cdot1+3x\cdot1^2+1^3+5\)
\(=\left(x+1\right)^3+5\)
Thay `x=19` vào \(\left(x+1\right)^3+5\) được:
\(\left(19+1\right)^3+5=8000+5=8005\)
b, \(x^3-3x^2+3x\)
\(=x^3-3x^2\cdot1+3x\cdot x-1^3+1\)
\(=\left(x-1\right)^3+1\)
Thay `x=11` vào \(\left(x-1\right)^3+1\) được:
\(\left(x-1\right)^3+1=\left(11-1\right)^3+1=10^3+1=1001\)
`a, (x+1)^3 + 5`
`= 20^3 + 5`
`= 805`
`b, (x-1)^3 - 1`
`= 10^3 - 1`
`= 999`