Question

27p^2-(1/216)-(9/2)p^ 2+(1/4)p

Answer 1

27p^3 - 1/216 - 9/2p^2 + 1/4p
= (3p)^3 - (1/6)^3 - 9/2p^2 + 1/4p
= (3p - 1/6)[(3p)^2 + p/2 + 1/36)] - (3p/2)(3p - 1/6)
= (3p - 1/6)(9p^2 + p/2 + 1/36 - 3p/2)
= (3p - 1/6)(9p^2 - 2p/2 + 1/36)
= (3p - 1/6)(9p^2 - p + 1/36)
= (3p - 1/6)(3p - 1/6)(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...☺