chứng minh
a3 +b3 +c3 - 3abc = (a+b+c) ( a2 + b2 + c2 - ab - bc - ca)
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\(=\left(7x-1-2x-1\right)\left(7x-1+2x+1\right)\)
\(=\left(5x-2\right)9x\)
neu co sai bn thong cam nha
a) \(2x^2+3x-5=2x^2-2x+5x-5\)
\(=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
b) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3z\left(x+y\right)\left(x+y+z\right)+z^3-x^3-y^3-z^3\)
\(=x^3+y^3+z^3+3xy\left(x+y\right)+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=3\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
c) \(x^4+x^3+x+1=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)=\left(x+1\right)^2\left(x^2-x+1\right)\)
d) \(x^4-x^3-x^2+1=x^2\left(x-1\right)-\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x^2-x-1\right)\)
e) \(x^4+4x^2-5=\left(x^2+2\right)^2-9=\left(x^2+2+3\right)\left(x^2+2-3\right)=\left(x^2+5\right)\left(x+1\right)\left(x-1\right)\)
\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(VT=a^3+b^3+c^3-3abc=\left(a+b\right)^3-3a^2b-3ab^2-3abc+c^2\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\right]\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\left(đpcm\right)\)