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\(\left(x+y\right)^3-\left(x^3+y^3\right)\)
\(=x^3+3x^2y+3xy^2+y^3-x^3-y^3\)
\(=3xy\left(x+y\right)\)
\((x+y)^3-(x-y)^3\)
\(=x^3+3x^2y+3xy^2+y^3-(x^3-3x^2y+3xy^2-y^3)\)
\(=6x^2y+2y^3\)
Cách khác:
Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\left(3x^2+y^2\right)\)
\(=6x^2y+2y^3\)
\(D=x^3-3x^2y+3xy^2-y^3-3x^3+6x^2y-3xy^2+3x^3-3x^2y-x^3\\ D=-y^3\)
Ta có: \(A=\left(x-y-1\right)^3-\left(x-y+1\right)^3+6\left(x-y\right)^2\)
\(=\left(x-y-1-x+y-1\right)\left[\left(x-y-1\right)^2+\left(x-y-1\right)\left(x-y+1\right)+\left(x-y+1\right)^2\right]+6\left(x-y\right)^2\)
\(=-2\cdot\left[3\left(x-y\right)^2+1\right]+6\left(x-y\right)^2\)
\(=-6\left(x-y\right)^2+6\left(x-y\right)^2-2\)
=-2
Q = (x - y)3 + (y + x)3 + (y - x)3 - 3xy(x + y)
= (x - y)3 - (x - y)3 + (x + y)[(x + y)2 - 3xy]
= (x + y)(x2 + 2xy + y2 - 3xy)
= (x + y)(x2 - xy + y2)
= x3 + y3
\(\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\)
\(=\left(x^4-y^4\right)\left(x^4+y^4\right)\)
\(=x^8-y^8\)
\(\left(x-y\right)^3+\left(x+y\right)^3\)
\(=x^3-3x^2y+3xy^2-y^3+x^3+3x^2y+3xy^2+y^3\)
\(=2x^3+6xy^2\)