\(a,=-\left(x^3-3x^2+3x-1\right)\\ =-\left(x-1\right)^3\\ b,=x^3+3.x^2\dfrac{1}{3}+3.\dfrac{1}{9}x+\left(\dfrac{1}{3}\right)^3\\ =\left(x+\dfrac{1}{3}\right)^3\)
\(a,-x^3+3x^2-3x+1=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(b,x^3+x^2+\dfrac{1}{3}x+\dfrac{1}{27}=\left(x-\dfrac{1}{3}\right)^2\)
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