\(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)

\(=a^3\left(b-c\right)-b^3.\left[b-c+a-b\right]+c^3\left(a-b\right)\)

\(=a^3\left(b-c\right)-b^3\left(b-c\right)-b^3\left(a-b\right)+c^3\left(a-b\right)\)

\(=\left(b-c\right)\left(a^3-b^3\right)-\left(a-b\right)\left(b^3-c^3\right)\)

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

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

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

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

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

Chúc bạn học tốt.

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

\(=a^3b-ab^3+b^3c-bc^3-ca\left(a^2-c^2\right)\)

\(=b\left(a^3-c^3\right)-b^3\left(a-c\right)-ca\left(a-c\right)\left(a+c\right)\)

\(=b\left(a-c\right)\left(a^2+ac+c^2\right)-b^3\left(a-c\right)-ca\left(a-c\right)\left(a+c\right)\)

\(=\left(a-c\right)\left(a^2b+abc+bc^2-b^3-a^2c-c^2a\right)\)

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

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

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

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

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

Vào câu hỏi tương tự đi

\(=x^4-x+2016x^2+2016x+2016.\)

\(=x\left(x^3-1\right)+2016\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2016\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2016\right)\)

Vào câu trả lời tương tự đi

a, \(x^{10}+x^5+1=(x^{10}-x)+(x^5-x^2)+(x^2+x+1)\)

\(=x(x^3-1)(x^6+x^3+1)+x^2(x^3-1)+(x^2+x+1)\)

\(=x(x-1)(x^2+x+1)(x^6+x^3+1)+x^2(x-1)(x^2+x+1)+(x^2+x+1)\)

Đến đây có nhân tử chung là \(x^2+x+1\) rồi bạn tự làm tiếp nha!

b, Tương tự câu a bạn cũng thêm bớt x và x^2

\(x^5+x^4+1=(x^5-x^2)+(x^4-x)+(x^2+x+1)\)

lọc cho ra nhân tử chung x^2+x+1 rồi giải tiếp

\(x^8+x+1=(x^8-x^2)+(x^2+x+1)\)

\(=x^2(x^6-1)+(x^2+x+1)=x^2(x^3-1)(x^3+1)+(x^2+x+1)\)

\(=x^2(x-1)(x^2+x+1)(x^3+1)+(x^2+x+1)\)

\(=(x^2+x+1)[x^2(x-1)(x^3+1)+1]\)

đến đây thôi! bạn tự xử tiếp nhé!!

\(P = 2a^3 + 7a^2b + 7ab^2 + 2b^3\)

\(=2a^3+2a^2b+5a^2b+5ab^2+2ab^2+2b^3\)

\(=2a^2(a+b)+5ab(a+b)+2b^2(a+b) \)

\(=(2a^2+5ab+2b^2)(a+b)\)

\(=(2a^2+4ab+ab+2b^2)(a+b)\)

\(=[2a(a+2b)+b(a+2b)](a+b)\)

\(=(2a+b)(2b+a)(a+b)\)

P=2a³+7a²b+7ab²+2b³

=2a³+2a²b+5a²b+5ab²+2ab²+2b²

=(2a³+2a²b)+(5a²b+5ab²)+(2ab²+2b³)

=2a²(a+b)+5ab(a+b)+2b²(a+b)

=(a+b)(2a²+5ab+2b²)

=(a+b)[2a²+4ab+ab+2b²]

=(a+b)[2a(a+2b)+b(a+2b)]

=(a+b)(2a+b)(a+2b)

Ta có :

\(x^2-y=y^2-x\)

\(\Rightarrow x^2-y^2 +x-y=0\)

\(\Rightarrow\left(x-y\right)\left(x+y+1\right)=0\)

Vì \(x\ne y\)

\(\Rightarrow x+y+1=0\)

\(\Rightarrow x+y=-1\)

Gọi biểu thức là  K . Ta có :

\(K=\left(x+y\right)^2-3\left(x+y\right)\)

\(=\left(-1\right)^2-3\left(-1\right)=4\)

Vậy ...

     \(x-y=xy-1\)

\(\Rightarrow xy-x+y-1=0\)

\(\Rightarrow x\left(y-1\right)+\left(y-1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(y-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\y=1\end{cases}}\)

TH1: Nếu \(x=-1\)thì \(x-y=-1-y\) và \(xy-1=-y-1\Rightarrow x-y=xy-1\)

TH2: Nếu y = 1 thì x - y = x - 1 và xy - 1 = x -1 nên x - 1 = xy - 1

Vậy x = -1 và y bất kỳ hoặc y = 1 và x bất kỳ.

Chúc bạn học tốt.

39^13+39^20

=39^13(39^7+1)

Có: 39^7+1 chia hết cho 40

=> 39^20+39^13 chia hết cho 40.

Ta có :

\(39^{20}+39^{13}\)

\(=39^{13}\left(39^7+1\right)⋮\left(39+1\right)=40\)

\(\Rightarrow39^{13}\left(39^7+1\right)⋮40\)

\(\Rightarrow39^{20}+39^{13}⋮40\) (đpcm)