`~rai~`
\(1/y^4-4x^2\\=(y^2)^2-(2x)^2\\=(y^2-2x)(y^2+2x).\\2/\left(x+\dfrac{1}{2}\right)^3\\=x^3+3x^2.\dfrac{1}{2}+3x.\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2} \right)^3\\=x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}.\\3/(2y+3x)^2\\=(2y)^2+2.2y.3x+(3x)^2\\=4y^2+12xy+9x^2.\\4/(3x-1)^3\\=(3x)^3-3.(3x)^2.1+3.3x.1^2-1^3\\=27x^3-27x^2+9x-1.\\5/\dfrac{y^4}{25}-16x^4\\=\left(\dfrac{y^2}{5}\right)-(4x^2)^2\\=\left(\dfrac{y^2}{5}-4x^2\right)\left(\dfrac{y^2}{5}+4x^2\right)\\=\left[\left(\dfrac{y}{\sqrt{5}}\right)^2-(2x)^2\right]\left(\dfrac{y^4}{5}+4x^2\right)\\=\left(\dfrac{y}{\sqrt{5}}-2x\right)\left(\dfrac{y}{\sqrt{5}}+2x\right)\left(\dfrac{y^2}{5}+4x^2\right)\)
