PT <=> \(x^2+8x+16+8x\left(x^2+8x+16\right)+15x^2\)

\(=x^2+8x+16+8x^3+64x^2+112+15x^2\)

\(=80x^2+8x^3+128=8\left(10x^2+x^3+16\right)\)

Số tự nhiên a chia 3 dư 2 => a có dạng 3k + 2 ( k ∈ N )

a² = ( 3k + 2 )² = 9k² + 12k + 4 = 9k² + 12k + 4

Ta có : 9k² chia hết cho 3 ; 12k chia hết cho 3 ; 4 chia 3 dư 1

=> 9k² + 12k + 4 chia 3 dư 1

hay a² chia 3 dư 1 ( đpcm )

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

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Vì \(\left(a-b\right)^2\ge0\), \(\left(b-c\right)^2\ge0\), \(\left(c-a\right)^2\ge0\)\(\forall a,b,c\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Leftrightarrow a=b=c\)

Vậy \(a=b=c\)( đpcm )

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

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Ta có : ( a - b )² ≥ 0 ∀ a, b ; ( b - c )² ≥ 0 ∀ b, c ; ( c - a )² ≥ 0 ∀ c, a

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)

Dấu "=" xảy ra khi và chỉ khi a = b = c ( đpcm )

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

\(=\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(2x^2+3\right)\)

2x⁴ - x² - 6 

= 2x⁴ - 4x² + 3x² - 6

= ( 2x⁴ - 4x² ) + ( 3x² - 6 )

= 2x²( x² - 2 ) + 3( x² - 2 )

= ( x² - 2 )( 2x² + 3 )

Ta có

  \(a^3+b^3+c^3-3abc=a^3+3a^2b+3ab^2+c^3-3abc-3a^2b-3ab^2\)

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

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

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

  Vậy  

 \(\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ac}=a+b+c\)