Question

factor 27p^3 -1/216-9/2p^2+1/4p
​

Answer 1

We have, 27p

3

−

216

1

−

2

9

p

2

+

4

1

p

Using, (a+b)

3

=a

3

+b

3

+3ab(a+b)

=(3p)

3

+(−

6

1

)

3

+3(3p)(−

6

1

)(3p−

6

1

)

=(3p−

6

1

)

3

=(3p−

6

1

)(3p−

6

1

)(3p−

6

1

) hope it will help you

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...☺