Skorzystamy ze wzorów skróconego mnożenia na sumę i różnicę sześcianów.

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$\text{a)}\left(x+2\right)\left(x^2-2x+4\right)=\left(x+2\right)\left(x^2-x\cdot 2+2^2\right)=x^3+2^3=x^3+8$

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$\text{b)}\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2+x\cdot 1+1^2\right)=x^3-1^3=x^3-1$

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$\text{c)}\left(x^2-x+1\right)\left(1+x\right)=\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left(x^2-x\cdot 1+1^2\right)=x^3+1^3=x^3+1$

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$\text{d)}\left(x^2-6x+36\right)\left(x+6\right)=\left(x+6\right)\left(x^2-6x+36\right)=\left(x+6\right)\left(x^2-x\cdot 6+6^2\right)=x^3+6^3=x^3+216$

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$\text{e)}\left(x-8\right)\left(x^2+8x+64\right)=\left(x-8\right)\left(x^2+x\cdot 8+8^2\right)=x^3-8^3=x^3-512$

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$\text{f)}\left(2x^2-1\right)\left(4x^4+2x^2+1\right)=\left(2x^2-1\right)\left({\left(2x^2\right)}^2+2x^2\cdot 1+1^2\right)={\left(2x^2\right)}^3-1^3=8x^6-1$