Lời giải:

Dựa vào các hằng đẳng thức đáng nhớ ta có:

\(a^3-3ab+2c=(x+y)^3-3(x+y)(x^2+y^2)+2(x^3+y^3)\)

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

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

\(=0\)

2/

2(x⁶+y⁶)-3(x⁴+y⁴)

=2[(x²)³+(y²)³ ] - 3x⁴-3y⁴

=2(x²+y²)(x⁴-x²y²+y⁴)-3x⁴-3y⁴

=2.1(x⁴-x²y²+y⁴)-3x⁴-3y⁴

=2x⁴-2x²y²+2y⁴-3x⁴-3y⁴

=-x⁴-2x²y²-y⁴

=-(x⁴+2x²y²+y⁴)

=-(x²+y²)

=-1

\(a^3-3ab+2c=0\)

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

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

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

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

\(=\left(x+y\right).0\)

\(=0\)

Sửa đề: Cho \(x+y=a;x^2+y^2=b;x^3+y^3=c\)

Chứng minh: \(a^3-2ab+2c=0\)

Giải:

Ta có:

\(a^3-3ab+2c=\left(x+y\right)^3-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)\)

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

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

\(=3\left(x+y\right)\left(x^2-xy+y^2+xy-x^2-y^2\right)=3\left(x+y\right).0\)

\(=0\) (đpcm)

Ta có \(a^3-3ab+2c=\left(x+y\right)^3-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)\)

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

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

\(=0\left(đpcm\right)\)

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

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

\(2c=2x^3+2y^3\)

\(a^3-3ab+2c=\left(x^3+y^3-3x^2-3y^2+2x^3+2y^3\right)+3\left(x^2y-xy^2+xy^2-xy^2\right)=0\)