\(\left(x+y\right)^2=4^2=16\\ =>\left(x^2+y^2\right)+2xy=16\\ =>12+2xy=16\\ =>2xy=16-12=4\\ =>xy=\dfrac{4}{2}=2\\ x^3-y^3\\ =\left(x-y\right)\left(x^2+xy+y^2\right)\\ =\sqrt{\left(x+y\right)^2-4xy}\cdot\left[\left(x^2+y^2\right)+xy\right]\\ =\sqrt{4^2-4\cdot2}\cdot\left(12+2\right)\\ =28\sqrt{2}\)
`x + y = 4`
`=> (x+y)^2 = 16`
`=> x^2 + 2xy + y^2 = 16`
`=> 12 + 2xy = 16`
`=> 2xy = 4`
`=> xy = 2`
Khi đó: `x^3 - y^3 = (x - y)(x^2 + xy + y^2) = (x-y) . (12 + 2) = 14 (x-y) `
Ta có: `x - y = sqrt{(x-y)^2} = sqrt{x^2 - 2xy + y^2} = sqrt{12 - 4} = sqrt{8} = 2sqrt{2} `
`=> x^3 - y^3 = 12 . 2 sqrt{2} = 28 sqrt{2} `
