Question
Factorize:$\frac{8}{27}\text{x}^3+1+\frac{4}{3}\text{x}^2+2\text{x}$
✓
Answer
$\frac{8}{27}\text{x}^3+1+\frac{4}{3}\text{x}^2+2\text{x}$$=\Big(\frac{2}{3}\text{x}\Big)^3+(1)^3+3\times\Big(\frac{2}{3}\text{x}\Big)^2\times1+3(1)^2\times\Big(\frac{2}{3}\text{x}\Big)$
$=\Big(\frac{2}{3}\text{x}+1\Big)^3$$\left[\because a^3+b^3+3 a^2 b+3 a b^2=(a+b)^3\right]$
$=\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)$
$\therefore\frac{8}{27}\text{x}^3+1+\frac{4}{3}\text{x}^2+2\text{x}$
$=\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)$
$=\Big(\frac{2}{3}\text{x}+1\Big)^3$$\left[\because a^3+b^3+3 a^2 b+3 a b^2=(a+b)^3\right]$
$=\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)$
$\therefore\frac{8}{27}\text{x}^3+1+\frac{4}{3}\text{x}^2+2\text{x}$
$=\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)$
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