Question

(3a+b)(9a2-3ab+b2)(27a3-b3)​

Answer 1

Answer:

$\mathrm{Expand\:\left(3a+b\right)\left(9a^2-3ab+b^2\right)\left(27a^3-b^3\right):\quad 729a^6+9a^2b^4-9b^4a^2-b^6}$

Step-by-step explanation:

$\mathrm{\left(3a+b\right)\left(9a^2-3ab+b^2\right)\left(27a^3-b^3\right)}$

$\mathrm{=\left(81a^4-3ab^3+27a^3b-b^4\right)\left(9a^2-3ab+b^2\right)}$

$\mathrm{Apply\:the\:distributive\:law:\quad \left(a+b\right)\left(x+y+z\right)=ax+ay+az+bx+by+bz}$

$\mathrm{\left(81a^4-3ab^3+27a^3b-b^4\right)\left(9a^2-3ab+b^2\right)=81a^4\cdot \:9a^2+81a^4\left(-3ab\right)+81a^4b^2-3ab^3\cdot}$$\mathrm{\:9a^2-3ab^3\left(-3ab\right)-\:9a^2-3ab^3\left(-3ab\right)-3ab^3b^2+27a^3b\cdot \:9a^2+27a^3b\left(-3ab\right)+}$

$\mathrm{27a^3bb^2-b^4\cdot \:9a^2-b^4\left(-3ab\right)-b^4b^2}$

$\mathrm{=81a^4\times9a^2+81a^4\left(-3ab\right)+81a^4b^2-3ab^3\cdot \:9a^2-3ab^3\left(-3ab\right)-3ab^3b^2+27a^3b}$$\mathrm{\times \:9a^2+27a^3b\left(-3ab\right)+27a^3bb^2-b^4\cdot \:9a^2-b^4\left(-3ab\right)-b^4b^2}$

$\mathrm{=729a^6+9a^2b^4-9b^4a^2-b^6}$