Bài 5:
\(a.x^3+3x^2+3x+2=0\\ < =>\left(x^3+3x^2+3x+1\right)+1=0\\ < =>\left(x+1\right)^3=-1\\ < =>\left(x+1\right)^3=\left(-1\right)^3\\ < =>x+1=-1\\ < =>x=-1-1\\ < =>x=-2\\ b.x^3-12x^2+48x-72=0\\ < =>\left(x^3-12x^2+48x-64\right)-8=0\\ < =>\left(x-4\right)^3=8\\ < =>\left(x-4\right)^3=2^3\\ < =>x-4=2\\ < =>x=2+4\\ < =>x=6\)
Bài 6:
Ta có:
\(a^3+b^3+3ab\\ =a^3+b^3+3ab\cdot1\\ =a^3+b^3+3ab\left(a+b\right)\\ =a^3+3a^2b+3ab^2+b^3\\ =\left(a+b\right)^3\\ =1^3\\ =1\)
Bài 4:
a: \(A=x^3+6x^2+12x+8\)
\(=x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3\)
\(=\left(x+2\right)^3\)
Khi x=8 thì \(A=\left(8+2\right)^3=10^3=1000\)
b: \(B=x^3-3x^2+3x-1\)
\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\)
\(=\left(x-1\right)^3\)
Khi x=101 thì \(B=\left(101-1\right)^3=100^3=1000000\)
c: \(D=2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)
\(=2\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]-3\left[\left(x+y\right)^2-2xy\right]\)
\(=2\left(1-3xy\right)-3\left(1-2xy\right)\)
=2-6xy-3+6xy=-1
