Question

Factorise 27p cube -1/216-9/2p square-1/4p

Answer 1

Answer -

Factorise :

          (27p³ - 1) / 216 - 9 / 2p² - 1 / 4p

                   ⇒ [ 216(9) - 2p² (27p³ - 1) ] / 216 × 2p² - [1 / 4p]

                   ⇒ [ 1944 - 54p^5 + 2p² ] / 432p² - [1 / 4p]

                  ⇒ 432p² (1) - 4p (1944 - 54p^5 + 2p²)

                  ⇒ 432p² - 7776p + 216p^6 - 8p³

               ⇒ 216p^6 - 8p³ + 432p² - 7776p

               

HOPE IT HELPS ...

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...☺