đơn giản, cứ áp dụng theo công thức là ra!!!!

\(a\left(b+c\right)^2\left(b-c\right)+b\left(c+a\right)^2\left(c-2\right)+c\left(a+b\right)^2\left(a-b\right)\)

\(=\left(b-c\right)\left(c-a\right)\left(c-b\right)\left(c+b+a\right)\)

nguồn câu hỏi tương tự

Trang 136 trong nâng cao phát triển có viết rồi mình cóp nó vô để mọi người dễ đọc nhé !

a) \(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)

\(=\left(x^4-2x^3+5x^2-4x+4\right)+\left(x^2-4x+4\right)\)

\(=x^4-2x^3+6x^2-8x+8\)

\(=\left(x^4-2x^3+2x^2\right)+\left(4x^2-8x+8\right)\)

\(=x^2\left(x^2-2x+2\right)+4\left(x^2-2x+2\right)\)

\(=\left(x^2+4\right)\left(x^2-2x+2\right)\)

\(x^4-9x^3+28x^2-36x+16\)

\(=x^4-x^3-8x^3+8x^2+20x^2-20x-16x+16\)

\(=\left(x^4-x^3\right)-\left(8x^3-8x^2\right)+\left(20x^2-20x\right)-\left(16x-16\right)\)

\(=x^3\left(x-1\right)-8x^2\left(x-1\right)+20x\left(x-1\right)-16\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3-8x^2+20x-16\right)\)

\(=\left(x-1\right)\left(x^3-2x^2-6x^2+12x+8x-16\right)\)

\(=\left(x-1\right)[x^2\left(x-2\right)-6x\left(x-2\right)+8\left(x-2\right)]\)

\(=\left(x-1\right)\left(x-2\right)\left(x^2-6x+8\right)\)

\(=\left(x-1\right)\left(x-2\right)\left(x^2-4x-2x+8\right)\)

\(=\left(x-1\right)\left(x-2\right)[x\left(x-4\right)-2\left(x-4\right)]\)

\(=\left(x-1\right)\left(x-2\right)\left(x-2\right)\left(x-4\right)\)

\(=\left(x-1\right)\left(x-2\right)^2\left(x-4\right)\)

Để tui cân hết choa 

a, Lời giải 

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

\(=x^4-2.x^2\left(x-2\right)+\left(x-2\right)^2+\left(x-2\right)^2\)

\(=x^4-2x^3+4x^2+2x^2-8x+8\)

\(=x^4-2x^3+6x^2+8x+8\)

\(=\left(x^4-2x^3+2x^2\right)+\left(4x^2-8x+8\right)\)

\(=x^2\left(x^2-2x+2\right)+4\left(x^2-2x+2\right)\)

\(=\left(x^2-2x+2\right)\left(x^2+4\right)\)

\(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)

\(=ab^2-ac^2+ca^2-cb^2+b\left(c^2-a^2\right)\)

\(=\left(ab^2-cb^2\right)-\left(ac^2-ca^2\right)+b\left(c-a\right)\left(c+a\right)\)

\(=b^2\left(a-c\right)-ac\left(c-a\right)+b\left(c-a\right)\left(c+a\right)\)

\(=b^2\left(a-c\right)+ac\left(a-c\right)-b\left(a-c\right)\left(c+a\right)\)

\(=\left(a-c\right)\left[b^2+ac-b\left(c+a\right)\right]\)

\(=\left(a-c\right)\left(b^2+ac-bc-ab\right)\)

\(=\left(a-c\right)\left[b\left(b-c\right)+a\left(c-b\right)\right]\)

\(=\left(a-c\right)\left[b\left(b-c\right)-a\left(b-c\right)\right]\)

\(=\left(a-c\right)\left(b-c\right)\left(b-a\right)\)

Cách khác:

Ta có:
\(a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)\)

\(=a(b^2-c^2)-b[(b^2-c^2)+(a^2-b^2)]+c(a^2-b^2)\)

\(=a(b^2-c^2)-b(b^2-c^2)-b(a^2-b^2)+c(a^2-b^2)\)

\(=(a-b)(b^2-c^2)-(b-c)(a^2-b^2)\)

\(=(a-b)(b-c)(b+c)-(b-c)(a-b)(a+b)\)

\(=(a-b)(b-c)[(b+c)-(a+b)]=(a-b)(b-c)(c-a)\)

Quy đồng đi, ta sẽ được  \(A=0\)

\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-c\right)\left(b-a\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)

\(A=\frac{-b+c}{-\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{-c+a}{-\left(a-b\right)\left(a-c\right)\left(b-a\right)}+\frac{-a+b}{-\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(A=\frac{-b+c-c+a-a+b}{-\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(A=\frac{0}{-\left(a-b\right)\left(a-c\right)\left(b-a\right)}\)

A = 0

Ta có:

a²(b + c) = b²(a + c)

<=> a² b - b² a + a² c - b² c = 0

<=> (a - b)(ab + bc + ca) = 0

<=> ab + bc + ca = 0 (vì a,b,c khác nhau từng đôi 1)

\(\Rightarrow\hept{\begin{cases}a\left(b+c\right)+bc=0\\c\left(a+b\right)+ab=0\end{cases}}\)

Ta lại có: a²(b + c) = 2016

<=> a(-bc) = 2016

<=> - abc = 2016

Ta xét 

P = c²(a + b) = c(-ab) = - abc = 2016

Không thấy ai tham gia nhỉ: Thảo luận cho vui nào?

\(\hept{\begin{cases}a^2\left(b+c\right)=2016\\b^2\left(a+c\right)=2016\\c^2\left(a+b\right)=2016\end{cases}\Rightarrow}\)có nghiệm không?

2016 nha