TeX Quellcode:

{\left(\dfrac{1}{4}x-1\right)^2 \left(8x+16\right) = \left(\dfrac{1}{16}x^2-\dfrac{1}{2}x+1\right)\left(8x+16\right) = \dfrac{1}{2}x^3-4x^2+8x+x^2-8x+16 = \dfrac{1}{2}x^3-3x^2+16}