Question

factorise 27 P cube - 1 / 216 - 9 / 2 p square + 1 / 4 P

Answer 1

hope this answer is helpful

Answer 2

Answer:

$27p^{3}-\frac{1}{216}-\frac{9p^{2}}{2}+\frac{p}{4}\\=\left( 3p - \frac{1}{6}\right)\left( 3p - \frac{1}{6}\right)\left( 3p - \frac{1}{6}\right)$

Step-by-step explanation:

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

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

$= \left( 3p - \frac{1}{6}\right)^{3}$

/* By algebraic identity :

x³ - 3x²y + 3xy² - y³ = (x-y)³ */

$=\left( 3p - \frac{1}{6}\right)\left( 3p - \frac{1}{6}\right)\left( 3p - \frac{1}{6}\right)$

Therefore.,

$27p^{3}-\frac{1}{216}-\frac{9p^{2}}{2}+\frac{p}{4}\\=\left( 3p - \frac{1}{6}\right)\left( 3p - \frac{1}{6}\right)\left( 3p - \frac{1}{6}\right)$

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