Từ hằng đẳng thức đã cho, ta có:

$2^3={\left(1+1\right)}^3=1^3+{3.1}^2+3.1+1$

$3^3={\left(2+1\right)}^3=2^3+{3.2}^2+3.2+1$

$4^3={\left(3+1\right)}^3=3^3+{3.3}^2+3.3+1$

...................................

${\left(n+1\right)}^3=n^3+3.n^2+3.n+1$

Cộng từng vế n đẳng thức trên rồi rút gọn, ta được ${\left(n+1\right)}^3=1^3+3\left(1^2+2^2+3^2+\mathrm{...}+n^2\right)+3\left(1+2+3+\mathrm{...}+n\right)+n$

Do đó

$3\left(1^2+2^2+3^2+\mathrm{...}+n^2\right)={\left(n+1\right)}^3-\frac{3n\left(n+1\right)}{2}-\left(n+1\right)$

$\begin{array}{l}3S=\left(n+1\right)\left[{\left(n+1\right)}^2-\frac{3n}{2}-1\right]\\ 3S=\left(n+1\right)\left(n^2+\frac{n}{2}\right)\\ 3S=\frac{1}{2}n\left(n+1\right)\left(2n+1\right)\end{array}$

Vậy $S=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)$

$$\begin{gathered} {\left(ab-1\right)}^2+{\left(a+b\right)}^2 \\ =a^2{b}^2-2ab+1+a^2+2ab+b^2 \end{gathered}$$

$\begin{array}{l}=\left(a^2{b}^2+a^2\right)+\left(b^2+1\right)\\ =a^2\left(b^2+1\right)+\left(b^2+1\right)\\ =\left(b^2+1\right)\left(a^2+1\right)\end{array}$

$$\begin{gathered} x^3+2x^2+2x+1 \\ =\left(x^3+1\right)+\left(2x^2+2x\right) \end{gathered}$$

$\begin{array}{l}=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\\ =\left(x+1\right)\left(x^2+x+1\right)\end{array}$

$$\begin{gathered} x^3-4x^2+12x-27 \\ =\left(x^3-27\right)-\left(4x^2-12x\right) \end{gathered}$$

$\begin{array}{l}=\left(x-3\right)\left(x^2+3x+9\right)-4x\left(x-3\right)\\ =\left(x-3\right)\left(x^2+3x+9-4x\right)\\ =\left(x-3\right)\left(x^2-x+9\right)\end{array}$

$$\begin{gathered} x^4-2x^3+2x-1 \\ =\left(x^4-1\right)-\left(2x^3-2x\right) \end{gathered}$$

$\begin{array}{l}=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\\ =\left(x^2-1\right)\left(x^2+1-2x\right)\\ =\left(x-1\right)\left(x+1\right){\left(x-1\right)}^2\\ =\left(x+1\right){\left(x-1\right)}^3\end{array}$

$$\begin{gathered} x^4+2x^3+2x^2+2x+1 \\ =\left(x^4+2x^3+x^2\right)+\left(x^2+2x+1\right) \end{gathered}$$

$\begin{array}{l}=x^2\left(x^2+2x+1\right)+{\left(x+1\right)}^2\\ =x^2{\left(x+1\right)}^2+{\left(x+1\right)}^2\\ ={\left(x+1\right)}^2\left(x^2+1\right)\end{array}$

$$\begin{gathered} x^2-2x-4y^2-4y \\ =\left(x^2-4y^2\right)-\left(2x+4y\right) \end{gathered}$$

$\begin{array}{l}=\left(x+2y\right)\left(x-2y\right)-2\left(x+2y\right)\\ =\left(x+2y\right)\left(x-2y-2\right)\end{array}$

$x^4+2x^3-4x-4$

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

$\begin{array}{l}=\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)\\ =\left(x^2-2\right)\left(x^2+2x+2\right)\end{array}$

$$\begin{gathered} x^2(1-x^2)-4-4x^2 \\ =x^2-x^4-4-4x^2 \end{gathered}$$

$\begin{array}{l}=x^2-{\left(x^2+2\right)}^2\\ =\left(x-x^2-2\right)\left(x+x^2+2\right)\end{array}$

$$\begin{gathered} (1+2x)(1-2x) - x(x+2)(x-2) \\ =1-4x^2-x\left(x^2-4\right) \end{gathered}$$

$\begin{array}{l}=1-4x^2-x^3+4x\\ =\left(1-x^3\right)-\left(4x^2-4x\right)\\ =\left(1-x\right)\left(1+x+x^2\right)+4x\left(1-x\right)\\ =\left(1-x\right)\left(x^2+5x+1\right)\end{array}$

$$\begin{gathered} x^2+y^2-x^2{y}^2+xy-x-y \\ =\left(x^2-x\right)+\left(y^2-x^2{y}^2\right)+\left(xy-y\right) \end{gathered}$$

$$\begin{gathered} \begin{array}{l}=x\left(x-1\right)+y^2\left(1-x^2\right)+y\left(x-1\right)\\ =x\left(x-1\right)+y^2\left(x-1\right)\left(x+1\right)+y\left(x-1\right)\end{array} \\ \begin{array}{l}=\left(x-1\right)\left[x-y^2\left(x+1\right)+y\right]\\ =\left(x-1\right)\left[\left(x-y^{2}x\right)-\left(y^2-y\right)\right]\\ =\left(x-1\right)\left[x\left(1-y^2\right)-y\left(y-1\right)\right]\\ =\left(x-1\right)\left[x\left(1-y\right)\left(1+y\right)+y\left(1-y\right)\right]\\ =\left(x-1\right)\left(1-y\right)\left(x+xy+y\right)\end{array} \end{gathered}$$

${199}^3-199=199.\left({199}^2-1\right)$

$\begin{array}{l}=199.\left(199+1\right).\left(199-1\right)\\ =\mathrm{198.199.200}⋮200\end{array}$

Vậy ${199}^3-199$ chia hết cho 200

$$\begin{gathered} A=x^6-2x^4+x^3+x^2-x \\ \begin{array}{l}=\left(x^6-2x^4+x^2\right)+\left(x^3-x\right)\\ ={\left(x^3-x\right)}^2+\left(x^3-x\right)\\ =6^2+6=42\end{array} \end{gathered}$$

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

$=a{b}^2+a{c}^2+abc+b{c}^2+b{a}^2+abc+c{a}^2+c{b}^2+abc$

$\begin{array}{l}=\left(a{b}^2+abc+b{a}^2\right)+\left(a{c}^2+abc+c{a}^2\right)+\left(b{c}^2+abc+c{b}^2\right)\\ =ab\left(b+c+a\right)+ac\left(c+b+a\right)+bc\left(c+a+b\right)\\ =\left(a+b+c\right)\left(ab+bc+ca\right)\end{array}$

$(a+b+c) (ab+bc+ca) -abc$

$=\left(a+b\right)\left(ab+bc+ca\right)+c\left(ab+bc+ca\right)-abc$

$=\left(a+b\right)\left(ab+bc+ca\right)+abc+c\left(bc+ca\right)-abc$

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

$\begin{array}{l}=\left(a+b\right)\left[\left(ab+ac\right)+\left(bc+c^2\right)\right]\\ =\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\\ =\left(a+b\right)\left(b+c\right)\left(c+a\right)\end{array}$

$$\begin{gathered} a(a+2b{)}^3-b(2a+b{)}^3 \\ =a{\left[\left(a+b\right)+b\right]}^3-b{\left[a+\left(a+b\right)\right]}^3 \end{gathered}$$

$=a\left[{\left(a+b\right)}^3+3b{\left(a+b\right)}^2+3b^2\left(a+b\right)+b^3\right]-b\left[a^3+3a^2\left(a+b\right)+3a{\left(a+b\right)}^2+{\left(a+b\right)}^3\right]$

$=a{\left(a+b\right)}^3+3ab{\left(a+b\right)}^2+3a{b}^2\left(a+b\right)+a{b}^3-b{a}^3-3b{a}^2\left(a+b\right)-3ab{\left(a+b\right)}^2-b{\left(a+b\right)}^3$

$=\left(a-b\right){\left(a+b\right)}^3+3ab\left(a+b\right)\left(b-a\right)+ab\left(b-a\right)\left(b+a\right)$

$=\left(a-b\right)\left(a+b\right)\left[{\left(a+b\right)}^2-3ab-ab\right]$

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

$=\left(a-b\right)\left(a+b\right){\left(a-b\right)}^2$

$=\left(a+b\right){\left(a-b\right)}^3$

$ab (a+b) -bc (b+c) +ac (a-c)$

$$\begin{gathered} =ab\left(a+b\right)-b^{2}c-b{c}^2+a^{2}c-a{c}^2 \\ =ab\left(a+b\right)+\left(a^{2}c-b^{2}c\right)-\left(a{c}^2+b{c}^2\right) \\ =ab\left(a+b\right)+c\left(a^2-b^2\right)-c^2\left(a+b\right) \\ =ab\left(a+b\right)+c\left(a-b\right)\left(a+b\right)-c^2\left(a+b\right) \\ =\left(a+b\right)\left(ab+ac-bc-c^2\right) \\ =\left(a+b\right)\left[\left(ab-bc\right)+\left(ac-c^2\right)\right] \\ =\left(a+b\right)\left[b\left(a-c\right)+c\left(a-c\right)\right] \\ =\left(a+b\right)\left(b+c\right)\left(a-c\right) \end{gathered}$$

$a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)+2abc$

$$\begin{gathered} =a{b}^2+a{c}^2+b{c}^2+b{a}^2+c\left(a^2+b^2+2ab\right) \\ =\left(a{b}^2+a^{2}b\right)+\left(a{c}^2+b{c}^2\right)+c{\left(a+b\right)}^2 \\ =ab\left(a+b\right)+c^2\left(a+b\right)+c{\left(a+b\right)}^2 \\ =\left(a+b\right)\left(ab+c^2+ac+bc\right) \\ =\left(a+b\right)\left[\left(ab+ac\right)+\left(bc+c^2\right)\right] \\ =\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right] \\ =\left(a+b\right)\left(b+c\right)\left(c+a\right) \end{gathered}$$

Nhận thấy $b^2-c^2=-\left[\left(a^2-b^2\right)+\left(c^2-a^2\right)\right]$ nên

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

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

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

$$\begin{gathered} =\left(a-c\right)\left(a^2-b^2\right)+\left(a-b\right)\left(c^2-a^2\right) \\ =\left(a-c\right)\left(a-b\right)\left(a+b\right)+\left(a-b\right)\left(c-a\right)\left(c+a\right) \\ =\left(a-b\right)\left[\left(a-c\right)\left(a+b\right)-\left(a-c\right)\left(c+a\right)\right] \\ =\left(a-b\right)\left(a-c\right)\left(b-c\right) \end{gathered}$$

Nhận thấy $c-a=-\left[\left(b-c\right)+\left(a-b\right)\right]$ nên

$$\begin{gathered} a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right) \\ =a^3\left(b-c\right)-b^3\left[\left(b-c\right)+\left(a-b\right)\right]+c^3\left(a-b\right) \\ =a^3\left(b-c\right)-b^3\left(b-c\right)-b^3\left(a-b\right)+c^3\left(a-b\right) \\ =\left(b-c\right)\left(a^3-b^3\right)-\left(a-b\right)\left(b^3-c^3\right) \\ =\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right) \\ =\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2-b^2-bc-c^2\right) \\ =\left(b-c\right)\left(a-b\right)\left[\left(a^2-c^2\right)+\left(ab-bc\right)\right] \\ =\left(b-c\right)\left(a-b\right)\left[\left(a-c\right)\left(a+c\right)+b\left(a-c\right)\right] \\ =\left(b-c\right)\left(a-b\right)\left(a-c\right)\left(a+b+c\right) \end{gathered}$$

Đặt $a+b-c=x,b+c-a=y,c+a-b=z$.

Khi đó $x+y+z=a+b-c+b+c-a+c+a-b=a+b+c$

Áp dụng hằng đẳng thức:

${\left(x+y+z\right)}^3-x^3-y^3-z^3=3\left(x+y\right)\left(y+z\right)\left(z+x\right)$

Ta có:

$$\begin{gathered} {\left(a+b+c\right)}^3-{\left(a+b-c\right)}^3-{\left(b+c-a\right)}^3-{\left(c+a-b\right)}^3 \\ =3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(c+a-b+a+b-c\right) \\ =3.2b.2c.2a \\ =24abc \end{gathered}$$

$$\begin{gathered} abc -(ab+bc+ca) + (a+b+c) -1 \\ =abc-ab-bc-ca+a+b+c-1 \\ =abc-bc-ab+b-ca+c+a-1 \\ =bc\left(a-1\right)-b\left(a-1\right)-c\left(a-1\right)+\left(a-1\right) \\ =\left(a-1\right)\left(bc-b-c+1\right) \\ =\left(a-1\right)\left[c\left(b-1\right)-\left(b-1\right)\right] \\ =\left(a-1\right)\left(b-1\right)\left(c-1\right) \end{gathered}$$

Vậy trong ba số a, b và c tồn tại hai số bằng nhau.

$$\begin{gathered} a^2+b^2=2ab\iff a^2-2ab+b^2=0 \\ \iff {\left(a-b\right)}^2=0\iff a-b=0\iff a=b \end{gathered}$$

Vậy $a^2+b^2=2ab$ thì $a=b$