Question

factorize : 27p^3-1/216-9p^2+1/4p​

Answer 1

Step-by-step explanation:

27p^3-1/216-9p^2+1/4

=(3p)^3-(1/6)^3-3×3p×p/2+1/4p

=(3p)^3-(1/6)^3-3×3p/6(3p-1/6)

=(3p-1/6)^3

Answer 2

Step-by-step explanation:

Given:-

$\tt{27 {p}^{3} - \frac{1}{216} - \frac{9}{2} {p}^{2} + \frac{1}{4} p}$

What to do:-

To Factorise the expression.

Solution:-

Let's solve the problem,

We have,

$\tt{27 {p}^{3} - \frac{1}{216} - \frac{9}{2} {p}^{2} + \frac{1}{4} p}$

$⟹ \tt{(3p {)}^{2} - \bigg( \frac{1}{6} \bigg )^{2} - \frac{3}{2} p \bigg(3p - \frac{1}{6} \bigg) }$

$⟹ \tt{\bigg(3p - \frac{1}{6 }\bigg)\left\{(3p {)}^{2} + 3p \times \frac{1}{6} + \bigg({ \frac{1}{6} \bigg)^{2} }\right\} - \frac{3}{2} p \bigg(3p - \frac{1}{6} \bigg) }\\ \\ </p><p>$

$⟹ \tt{\bigg(3p - \frac{1}{6} \bigg) \bigg(9 {p}^{2} + p \times \frac{1}{2} + \frac{1}{36} \bigg) - \frac{3}{2} p \bigg(3p - \frac{1}{6} \bigg) }$

$⟹ \tt{\bigg(3p - \frac{1}{6} \bigg) \bigg(9 {p}^{2} + \frac{p}{2} + \frac{1}{36} - \frac{3}{2} p \bigg)}$

$⟹ \tt{\bigg(3p - \frac{1}{6} \bigg) \bigg(9 {p}^{2} - p + \frac{1}{36} \bigg)}$

$⟹ \tt{\bigg(3p - \frac{1}{6} \bigg) \left \{(3p {)}^{2} - 2 \times 3p \times \frac{1}{6} + \bigg( \frac{1}{6} \bigg)^{2} \right \} }$

$⟹ \tt{\bigg(3p - \frac{1}{6} \bigg) \bigg(3p - \frac{1}{6} \bigg )^{2} }$

$⟹ \tt{\bigg(3p - \frac{1}{6} \bigg) \bigg(3p - \frac{1}{6} \bigg) \bigg(3p - \frac{1}{6} \bigg) } \: \: \red{Ans}.$

I hope it's help you...☺