a) \(14\left(x-y\right)^2+21\left(y-x\right)\)

\(=14\left(x-y\right)^2-21\left(x-y\right)\)

\(=7\left(x-y\right)\left[2\left(x-y\right)-3\right]\)

\(=7\left(x-y\right)\left(2x-2y-3\right)\)

b) \(7x^5\left(y-3\right)-49x^4\left(3-y\right)^3\)

\(=7x^4\left(y-3\right)\left[x+7\left(y-3\right)^2\right]\)

\(=7x^4\left(y-3\right)\left(x+7y^2-42y+63\right)\)

c) \(\left(x^2-9\right)^2-x^2\left(x-3\right)^2\)

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

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

\(=\left(x-3\right)^2\left(x^2+6x+9-x^2\right)\)

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

d) \(\left(4x^2-1\right)^2-9\left(2x-1\right)^2\)

\(=\left(2x-1\right)^2\left(2x+1\right)^2-9\left(2x-1\right)^2\)

\(=\left(2x-1\right)^2\left[\left(2x+1\right)^2-9\right]\)

\(=\left(2x-1\right)^2\left(4x^2+4x+1-9\right)\)

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

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

Ta có: \(\left(x+3\right)^3-\left(x+1\right)^3=56\)

\(\Leftrightarrow x^3+9x^2+27x+27-x^3-3x^2-3x-1-56=0\)

\(\Leftrightarrow6x^2+24x-30=0\)

\(\Leftrightarrow x^2+4x-5=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

\(\left(3x-2\right)^3=0\)

\(\left(3x-2\right)^3=0^3\)

\(\Rightarrow3x-2=0\)

\(\Rightarrow3x=2\)

\(\Rightarrow x=\frac{2}{3}\)

( 3x - 2 )³ = 0

<=> 3x - 2 = 0

<=> 3x = 2

<=> x =\(\frac{2}{3}\)

\(\left(\frac{1}{3}-2x\right)\left(4x^2+\frac{2}{3}x+\frac{1}{9}\right)-\left(\frac{1}{27}-8x^3\right)\)

\(=\frac{1}{3}\left(4x^2+\frac{2}{3}x+\frac{1}{9}\right)-2x\left(4x^2+\frac{2}{3}x+\frac{1}{9}\right)-\frac{1}{27}+8x^3\)

\(=\frac{4}{3}x^2+\frac{2}{9}x+\frac{1}{27}-8x^3-\frac{4}{3}x^2-\frac{2}{9}x-\frac{1}{27}+8x^3\)

\(=\left(\frac{4}{3}x^2-\frac{4}{3}x^2\right)+\left(\frac{2}{9}x-\frac{2}{9}x\right)+\left(\frac{1}{27}-\frac{1}{27}\right)+\left(-8x^3+8x^3\right)\)

= 0 =>không phụ thuộc vào biến x

Ta có: \(\left(\frac{1}{3}-2x\right)\left(4x^2+\frac{2}{3}x+\frac{1}{9}\right)-\left(\frac{1}{27}-8x^3\right)\)

\(=\left(\frac{1}{3}-2x\right)\left[\left(\frac{1}{3}\right)^2+\frac{1}{3}\cdot2x+\left(2x\right)^2\right]-\left(\frac{1}{27}-8x^3\right)\)

\(=\left(\frac{1}{27}-8x^3\right)-\left(\frac{1}{27}-8x^3\right)\)

\(=0\)

=> đpcm

Ta có (x - 1)³ - (x + 1)³ + 6(x + 1)(x - 1)

= x³ - 3x² + 3x - 1 - (x³ + 3x² + 3x + 1) + 6(x² - 1)

= x³ - 3x² + 3x - 1 - x³ - 3x² - 3x - 1 + 6x² - 6

= -6x² - 2 + 6x² - 6

= -8 

=> Biểu thức trên không phụ thuộc vào biến (đpcm)

Ta có: \(\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)

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

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

\(=-8\)

=> đpcm

( x + 2 )³ + ( x - 2 )³ - 2x( x² + 12 )

= x³ + 6x² + 12x + 8 + x³ - 6x² + 12x - 8 - 2x³ - 24x

= 0 ( đpcm )

\(2^{2^{6n+2}}+13⋮29\)

\(\Leftrightarrow4^{6n+2}+13⋮29\)

\(\Leftrightarrow16^{3n+1}+13⋮29\)

\(\Leftrightarrow\left(16+13\right)\left(3^n....+1\right)⋮29\left(dpcm\right)\)