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Phân tích mẫu thức thành nhân tử :
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2c-ab^2+ac^2-bc^2\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)\)
\(=\left(b-c\right)\left(a^2+bc-ab-ac\right)\)
\(=\left(b-c\right)\left[a\left(a-b\right)-c\left(a-b\right)\right]=\left(b-c\right)\left(a-c\right)\left(a-b\right).\)
Do đó : \(A=\frac{\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3}{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Nhận xét : Nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz.\)
Đặt \(b-c=x,c-a=y,a-b=z\) thì \(x+y+z=0\)
Theo nhận xét trên : \(A=\frac{x^3+y^3+z^3}{-xyz}=\frac{3xyz}{-xyz}=-3.\)
Tử:
(b - c)3 + (c - a)3 + (a - b)3
= (b - c + c - a + a - b)3 - 3(b - c + c - a)(b - c + a - b)(c - a + a - b)
= 0 - 3(b - a)(a - c)(c - b)
= 3(a - b)(a - c)(c - b)
Mẫu:
a2(b - c) + b2(c - a) + c2(a - b)
= a2(b - c) + b2c - ab2 + ac2 - bc2
= a2(b - c) - a(b2 - c2) + bc(b - c)
= a2(b - c) - a(b - c)(b + c) + bc(b - c)
= (b - c)(a2 - ab - ac + bc)
= (b - c)[a(a - b) - c(a - b)]
= (b - c)(a - b)(a - c)
\(A=\frac{3\left(a-b\right)\left(a-c\right)\left(c-b\right)}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}\)
\(=\frac{3\left(c-b\right)}{b-c}\)
Phân tích mẫu \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2c-ab^2+c^2a-c^2b\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b+c\right)\left(b-c\right)\)
\(=\left(b-c\right)\left(a^2+bc-ab-ac\right)=\left(b-c\right)\left[a\left(a-c\right)-b\left(a-c\right)\right]\)
\(=\left(b-c\right)\left(a-b\right)\left(a-c\right)=-\left(b-c\right)\left(a-b\right)\left(c-a\right)\)
Đặt b - c = x, c - a = y, a - b = z
=> x + y + z = b - c + c - a + a - b = 0
Từ x+y+z=0 => x3+y3+z3=3xyz (tự c/m)
=>\(A=\frac{x^3+y^3+z^3}{-xyz}=\frac{3xyz}{-xyz}=-3\)
\(=\dfrac{b\left(b-c\right)-a\left(a-c\right)}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{b^2-bc-a^2+ac}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{-\left(a-b\right)\left(a+b\right)+c\left(a-b\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{-a-b+c}{ab\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{1}{a\left(a-b\right)\left(a-c\right)}-\dfrac{1}{b\left(a-b\right)\left(b-c\right)}\)
\(=\dfrac{b^2-cb-a^2+ac}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(b-a\right)\left(b+a\right)-c\left(b-a\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{-\left(b+a-c\right)}{ab\left(a-c\right)\left(b-c\right)}\)
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}\)
\(=\dfrac{1}{a\left(a-b\right)\left(a-c\right)}-\dfrac{1}{b\left(a-b\right)\left(b-c\right)}\)
\(=\dfrac{b\left(b-c\right)-a\left(a-c\right)}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{b^2-bc-a^2+ac}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{-\left(a-b\right)\left(a+b\right)+c\left(a-b\right)}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{\left(a-b\right)\left(-a-b+c\right)}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=-\dfrac{a+b-c}{ab\left(b-c\right)\left(a-c\right)}\)