Math, asked by swtshreya5590, 10 months ago

27pcube-1÷216-9÷2p square +1÷4p

Answers

Answered by himanshubani1711
3

Answer:

Step-by-step explanation:

Using the identity

(a-b)^3=a^3-b^3-3a^2b+3ab^2

Attachments:
Answered by Salmonpanna2022
5

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...☺

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