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