تدرب وحل التمرينات

(1)- حلل كل مقدار من المقادير الجبرية الآتية إلى أبسط صورة:

$6^3+x^3$

$6^3+x^3=(6+x)(36-6x+x^2)$

$27+64x^3$

$27+64x^3=(3+4x)(9-12x+16x^2)$

$125y^3+1$

$125y^3+1=(5y+1)(25y^2-5y+1)$

$\frac{1}{64}+\frac{8}{125}{y}^3$

$\frac{1}{64}+\frac{8}{125}{y}^3=(\frac{1}{4}+\frac{2}{5}y)(\frac{1}{16}-\frac{1}{10}y+\frac{4}{25}{y}^2)$

$\frac{1}{b^3}+\frac{1}{8}$

$\frac{1}{h^3}+\frac{1}{8}=(\frac{1}{b}+\frac{1}{2})(\frac{1}{b^2}-\frac{1}{2b}+\frac{1}{4})$

$\frac{1}{5}{\mathrm{v}}^3+25$

$\frac{1}{5}{v}^3+25=\frac{1}{5}(v^3+125)=\frac{1}{5}(v+5)(v^2-5v+25)$

$0.027+27n^3$

$0.027+27n^3=(0.3+3n)(0.09-0.9n+9n^2)$

$0.125x^3+0.008y^3$

$0.125x^3+0.008y^3=(0.5x+0.2y)(0.25x^2-0.1xy+0.04y^2)$

(2)- حلل كل مقدار من المقادير الجبرية الآتية إلى أبسط صورة:

$y^3-64$

$y^3-64=(y-4)(y^2+4y+16)$

$27y^3-8$

$27y^3-8=(3y-2)(9y^2+6y+4)$

$\frac{1}{x^3}-\frac{27}{8}$

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

$9-\frac{1}{3}{n}^3$

$9-\frac{1}{3}{n}^3=\frac{1}{3}(27-n^3)=\frac{1}{3}(3-n)(9+3n+n^2)$

$0.001-{\mathrm{v}}^3$

$0.001-v^3=(0.1-v)(0.01v+0.1v+v^2)$

$4-\frac{1}{2}{t}^3$

$4-\frac{1}{2}{t}^3=\frac{1}{2}(8-t^3)=\frac{1}{2}(2-t)(4+2t+t^2)$

$25c^3-\frac{1}{5}$

$25c^3-\frac{1}{5}=\frac{1}{5}(125c^3-1)=\frac{1}{5}(5c-1)(25c^2+5c+1)$

$0.001x^3-0.008y^3$

$0.001x^3-0.008y^3=(0.1x-0.2y)(0.01x^2+0.02xy+0.04y^2)$