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

\(\left(36x^2-25\right)-\left(6x+5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(6x+5\right)\left(6x-5\right)-\left(6x+5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(6x+5\right)\left(6x-5-x-1\right)=0\)

\(\Leftrightarrow\left(6x+5\right)\left(5x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{6}\\x=\frac{6}{5}\end{cases}}\)

\(\left(36x^2-25\right)-\left(6x+5\right)\left(x+1\right)=0\Leftrightarrow\left(6x-5\right)\left(6x+5\right)-\left(6x+5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(6x+5\right)\left(5x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{6}\\x=\frac{6}{5}\end{cases}}\)

\(\left(2n+3\right)^2-25=\left(2n+3\right)^2-5^2\)

\(=\left(2n-2\right)\left(2n+8\right)\)

\(=4\left(n-1\right)\left(n+2\right)\)

Vì \(\left(n-1\right)\left(n+2\right)⋮2\)nên \(=4\left(n-1\right)\left(n+2\right)⋮8\)

\(\Rightarrow\left(2n+3\right)^2-25⋮8\)

a) \(3x^2-3y^2+2x+2y\)

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

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

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

d) \(4x^2-4x-3\)

\(=\left(2x\right)^2-2.2x+1-4\)

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

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

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