Question:
Find the value
$\frac{8}{27} x^{3}+1+\frac{4}{3} x^{2}+2 x$
Solution:
$=\left(\frac{2}{3} x^{3}\right)^{3}+1^{3}+3 \times\left(\frac{2}{3} x\right)^{2} \times 1+3(1)^{2} \times\left(\frac{2}{3} x\right)$
$=\left(\frac{2}{3} x+1\right)^{3}$
$\left[\therefore x^{3}+b^{3}+3 x^{2} b+3 x b^{2}=(x+b)^{3}\right]$
$=\left(\frac{2}{3} x+1\right)\left(\frac{2}{3} x+1\right)\left(\frac{2}{3} x+1\right)$
$\therefore \frac{8}{27} x^{3}+1+\frac{4}{3} x^{2}+2 x$
$=\left(\frac{2}{3} x+1\right)\left(\frac{2}{3} x+1\right)\left(\frac{2}{3} x+1\right)$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.