Question

Factorise the Following -
$27 {p}^{3} - \frac{1}{216} - \frac{9}{2} \: {p}^{2} + \frac{1}{4} \: p$
CLUE -
$use \: the \: identity \: > (x - y) {}^{3} = {x}^{3} - {y}^{3} - 3xy \: (x - y)$
Give a Correct Step-by-step Explanation:

REFERENCE -
Class 9 - NCERT - Mathematics - Polynomials - EXERCISE 2.5 - Main. 8 - (v)

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Answer 1

Answer:

27p³-1/216-9/2p²+1/4p

(3p)³ - (1/6)³ - 3×3p×1/6×3p + 3×3p×1/6×1/6

(3p-1/6)³

Step-by-step explanation:

By using the identity

(x-y)³=x³-y³-3xy(x-y)

=x³-y³-3x²y+3xy²

Hopefully it's helpful.

Answer 2

GIVEN :-

• 27 p³ - 1/216 - 9/2 p² + 1/4 p.
• (x - y)³ = x³ - y³ - 3x²y + 3xy².

TO FACTORISE :-

• 27 p³ - 1/216 - 9/2 p² + 1/4 p.

FACTORISATION :-

$\\ : \implies \displaystyle \sf \: 27p ^{3} - \dfrac{1}{216} - \dfrac{9}{2} p ^{2} + \dfrac{1}{4} p$

$: \implies \displaystyle \sf \:(3p) ^{3} - \bigg( \dfrac{1}{6} \bigg) ^{3} - 3 \times (3p) ^{2} \times \dfrac{1}{6} + 3 \times 3p \times \bigg( \dfrac{1}{6} \bigg) ^{2}$

$\bigstar \displaystyle \sf \:identity \to (x - y) ^{3} = x ^{3} - y^{3} - 3x ^{2} y + 3xy ^{2}$

$: \implies \displaystyle \sf \: \bigg(3p - \dfrac{1}{6} \bigg) ^{3}$

$: \implies \displaystyle \sf \:\bigg(3p - \dfrac{1}{6} \bigg)\bigg(3p - \dfrac{1}{6} \bigg)\bigg(3p - \dfrac{1}{6} \bigg)$

$\therefore \underline{\displaystyle \sf required\ Answer \ is \ \bigg(3p - \dfrac{1}{6} \bigg)^{3}}$