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\(\left(a+b\right)\left(a^2-b^2\right)+\left(b+c\right)\left(b^2-c^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)
\(=a^3-ab^2+a^2b-b^3+b^3-bc^2+b^2c-c^3+c^3-a^2c+ac^2-a^3\)
\(=-ab^2+a^2b-bc^2+b^2c-a^2c+ac^2\)
\(=\left(a^2b-ab^2\right)+\left(ac^2-bc^2\right)-\left(a^2c-b^2c\right)\)
\(=ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(ab+c^2-ac-bc\right)\)
\(=\left(a-b\right)\left[\left(ab-ac\right)+\left(c^2-bc\right)\right]\)
\(=\left(a-b\right)\left[a\left(b-c\right)+c\left(c-b\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
\(a^2b^2\left(a-b\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)-b^2c^2\left[\left(a-b\right)+\left(c-a\right)\right]+c^2a^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)-b^2c^2\left(a-b\right)+c^2a^2\left(c-a\right)-b^2c^2\left(c-a\right)\)
\(=\left(a-b\right)b^2\left(a-c\right)\left(a+c\right)+\left(c-a\right)c^2\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(ab^2+cb^2-c^2a-c^2b\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(ab+ac+bc\right)\)
1.
\(y^2+y\left(x^3+x^2+x\right)+x^5-x^4+2x^3-2x^2\)
\(\Delta=\left(x^3+x^2+x\right)^2-4\left(x^5-x^4+2x^3-2x^2\right)\)
\(=\left(x^3-x^2+3x\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}y=\dfrac{-x^3-x^2-x+x^3-x^2+3x}{2}=-x^2+x\\y=\dfrac{-x^3-x^2-x-x^3+x^2-3x}{2}=-x^3-2x\end{matrix}\right.\)
Hay đa thức trên có thể phân tích thành:
\(\left(x^2-x+y\right)\left(x^3+2x+y\right)\)
Dựa vào đó em tự tách cho phù hợp
#)Giải :
a)\(ab\left(b-a\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(=a\left(a-b\right)+b^2c-bc^2+ac^2-a^2c\)
\(=ab\left(a-b\right)-\left(a-b\right)\left(a+b\right)c+c^2\left(a-b\right)\)
\(=\left(ab-ac-bc+c^2\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
b) \(a^2\left(b-c\right)-b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(a^2-b^2\right)\left(b-c\right)-\left(b^2-c^2\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(b-c\right)-\left(b-c\right)\left(b+c\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
\(=2\left(x^2+x-5\right)^2-5\left(x^2+x-5\right)+3\)
\(=2\left(x^2+x-5\right)-2\left(x^2+x-5\right)-3\left(x^2+x-5\right)+3\)
\(=2\left(x^2+x-5\right)\left(x^2+x-6\right)-3\left(x^2+x-6\right)\)
\(=\left(x^2+x-6\right)\left(2x^2+2x-13\right)\)
\(=\left(x-2\right)\left(x+3\right)\left(2x^2+2x-13\right)\)
\(C=2\left(x^2+x-5\right)^2-5\left(x^2+x\right)+28\)
Đặt t=\(x^2+x\)
\(\Rightarrow C=2\left(t-5\right)^2-5t+28=2t^2-20t+50-5t+28=2t^2-25t+78=2\left(t-\dfrac{13}{2}\right)\left(t-6\right)\)
Thay t: \(C=2\left(t-\dfrac{13}{2}\right)\left(t-6\right)=2\left(x^2+x-\dfrac{13}{2}\right)\left(x^2+x-6\right)=2\left(x-2\right)\left(x+3\right)\left(x^2+x-\dfrac{13}{2}\right)\)
\(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)=\left(c-a\right)\left(c-b\right)\left(b-a\right)\)