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a) \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2y-x^2z+y^2\left(z-x\right)+z^2x-z^2y\)
\(=\left(x^2y-z^2y\right)+\left(z^2x-x^2z\right)+y^2\left(z-x\right)\)
\(=y\left(x+z\right)\left(x-z\right)-xz\left(x-z\right)-y^2\left(x-z\right)\)
\(=\left(x-z\right)\left(xy+yz-xz-y^2\right)\)
\(=\left(x-z\right)\left[\left(xy-xz\right)+\left(yz-y^2\right)\right]\)
\(=\left(x-z\right)\left[x\left(y-z\right)-y\left(y-z\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(y-z\right)\)
Em tham khảo:
Câu hỏi của Thanh Triều Vương Nguyễn - Toán lớp 8 - Học toán với OnlineMath
a) = a3+b3+c3 +3a2b +3ab2 -3ab(a+b) - 3abc
= (a+b)3+c3-3ab(a+b)-3abc (áp dụng A3+B3 ta có)
=(a+b+c) ( (a+b)2 - (a+b)c +c2) - 3ab(a+b+c)
=(a+b+c) ( (a+b)2 - (a+b)c +c2 - 3ab) (nhân tử chung là a+b+c)
=(a+b+c) ( a2+2ab+b2- ac-bc +c2 -3ab)
=(a+b+c) (a2+b2+c2-ab-ac-bc)
Phần b tương tự
a) \(a^3+b^3-c^3+3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)-c^3+3abc\)
\(=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]-3ab\left(a+b-c\right)\)
\(=\left(a+b-c\right)\left(a^2+2ab+b^2+ca+bc+c^2-3ab\right)\)
\(=\left(a+b-c\right)\left(a^2+b^2+c^2-ab+bc+ca\right)\)
b) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(x+y+z-x\right)\left[\left(x+y+z\right)^2+\left(x+y+z\right)x+x^2\right]-\left(y+z\right)\left(y^2-yz+z^2\right)\)
\(=\left(y+z\right)\left[x^2+y^2+z^2+2\left(xy+yz+zx\right)+x^2+xy+zx+x^2-y^2+yz-z^2\right]\)
\(=\left(y+z\right)\left(3x^2+3xy+3yz+3zx\right)\)
\(=3\left(y+z\right)\left[x\left(x+y\right)++z\left(x+y\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(a)a^3+b^3-c^3+3abc=\left(a+b\right)^3-3ab\left(a+b\right)-c^3+3abc\)
\(=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]-3ab\left(a+b-c\right)\)
\(=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab+bc+ac\right)\)