Find the value

Question:
Find the value

$\frac{8}{27} x^{3}+1+\frac{4}{3} x^{2}+2 x$

Solution:

$=\left(\frac{2}{3} x^{3}\right)^{3}+1^{3}+3 \times\left(\frac{2}{3} x\right)^{2} \times 1+3(1)^{2} \times\left(\frac{2}{3} x\right)$

$=\left(\frac{2}{3} x+1\right)^{3}$

$\left[\therefore x^{3}+b^{3}+3 x^{2} b+3 x b^{2}=(x+b)^{3}\right]$

$=\left(\frac{2}{3} x+1\right)\left(\frac{2}{3} x+1\right)\left(\frac{2}{3} x+1\right)$

$\therefore \frac{8}{27} x^{3}+1+\frac{4}{3} x^{2}+2 x$

$=\left(\frac{2}{3} x+1\right)\left(\frac{2}{3} x+1\right)\left(\frac{2}{3} x+1\right)$