\(27x^3-54x^2y+36xy^2-8y^3\)
\(=\left(3x-2y\right)^3\)
Tại \(x=4;\)\(y=6\) thì gtbt là:
\(\left(3.4-2.6\right)^3=0\)
\(27x^3-54x^2y+36xy^2-8y^3\)
\(=\left(3x\right)^3-3.\left(3x\right)^2.2y+3.\left(2y\right)^2.3x-\left(2y\right)^3\)
\(=\left(3x-2y\right)^3=\left(4.3-2.6\right)^3=0^3=0\)
\(27x^3-54x^2y+36xy^2-8y^3\)
\(=\left(3x-2y\right)^3\)
THAY x=4, y=6 ta có:
\(\left(3.4-2.6\right)^3\)
\(=0\)
\(27x^3-54x^2y+36xy^2-8y^3\)
\(=\left(3x-2y\right)^3\)
\(=\left(3\cdot4-2\cdot6\right)^3=0\)
Ta có : \(27x^3-54x^2y+36xy^2-8y^3=\left(3x-2y\right)^3\)
Thay x = 4 ; y = 6 ta có : \(\left(3.4-2.6\right)^3=\left(12-12\right)^3=0\)