Question

P=(3x +y)^{3 -}(2x-y) (2x +y) +(x-3y)³ tại x=1/3 : y= - 1/3

Answer 1

\(P=\left(3x+y\right)^3-\left(2x-y\right)+\left(x-3y\right)^3\)

    \(=3x^3+3.3x^2.y+3.3x.y^2+y^3\)\(-2x^2-y^2\)\(+x^3-3.x^2.3y+3.x.3y^2-y^3\)

    \(=\left(3x^3+x^3\right)\)\(+\left(9x^2y-9x^2y\right)\)\(+\left(9xy^2-9xy^2\right)\)\(+\left(y^3-y^3\right)\)\(-2x^2-y\)

   = \(4x^3-2x^2-y^2\)

Thay x=\(\dfrac{1}{3},y=-\dfrac{1}{3}\)

\(4.\left(\dfrac{1}{3}\right)^3-2.\left(\dfrac{1}{3}\right)^2-\left(\dfrac{-1}{3}\right)^2\)

=\(4.\dfrac{1}{27}-2.\dfrac{1}{9}-\dfrac{1}{9}=\dfrac{4}{27}-\dfrac{2}{9}-\dfrac{1}{9}=\dfrac{-5}{9}\)