2A:
a: \(x^3+12x^2+48x+64=\left(x+2\right)^3\)
b: \(-8x^3+12x^2+1-6x=\left(-2x+1\right)^3\)
c: \(\dfrac{8}{27}x^3-\dfrac{8}{3}x^2y+8x^2y-8y^3=\left(\dfrac{2}{3}x-2y\right)^3\)
d: \(\left(x-y\right)^6-6\left(x-y\right)^4+12\left(x-y\right)^2-8=\left[\left(x-y\right)^2-2\right]^3\)
1B:
a: Ta có: \(\left(x+3\right)^3\)
\(=x^3+3\cdot x^2\cdot3+3\cdot x\cdot3^2+3^3\)
\(=x^3+9x^2+27x+27\)
b: Ta có: \(\left(\dfrac{1}{2}-x\right)^3\)
\(=\left(\dfrac{1}{2}\right)^3-3\cdot\left(\dfrac{1}{2}\right)^2\cdot x+3\cdot\dfrac{1}{2}\cdot x^2-x^3\)
\(=\dfrac{1}{8}-\dfrac{3}{4}x+\dfrac{3}{2}x^2-x^3\)
c: Ta có: \(\left(2x-y^2\right)^3\)
\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot y^2+3\cdot2x\cdot y^4-y^6\)
\(=8x^3-12x^2y^2+6xy^4-y^6\)
d: Ta có: \(\left(2x-\dfrac{y^2}{x}\right)^3\)
\(=\left(2x\right)^3-3\cdot4x^2\cdot\dfrac{y^2}{x}+3\cdot2x\cdot\dfrac{y^4}{x^2}-\left(\dfrac{y^2}{x}\right)^3\)
\(=8x^3-12xy^2+\dfrac{6y^4}{x}-\dfrac{y^6}{x^3}\)
1A:
a: \(\left(3x+1\right)^3=27x^3+27x^2+9x+1\)
b: \(\left(\dfrac{1}{3}x-1\right)^3=\dfrac{1}{27}x^3-\dfrac{1}{3}x+x-1\)
c: \(\left(-y^2+3x\right)^3=27x^6-27x^2y^2+9xy^4-y^6\)
d: \(\left(2x-\dfrac{y^2}{x}\right)^3=8x^3-12xy^2+6y^2-\dfrac{y^6}{x^3}\)
