Question

Please factorise, no spam boys
$27p {}^{3} - \frac{1}{216} - \frac{9}{2} p {}^{2} + \frac{1}{4} p$​

Answer 1

Answer:

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

Some useful identities:

→ (a + b)^2 = a^2 + 2ab + b^2

→ (a – b)^2 = a^2 – 2ab + b^2

→ a^2 – b^2 = (a + b) (a – b)

→ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

→ (a + b – c)^2 = a^2 + b^2 + c^2 + 2ab – 2bc – 2ca

→ (a – b – c)^2 = a^2 + b^2 + c^2 – 2ab + 2bc – 2ca

→ (a + b)^3 = a^3 + b^3 + 3ab(a + b)

→ (a – b)^3 = a^3 – b^3 – 3ab(a – b)

→ (a^3 + b^3) = (a + b) (a^2 – ab + b^2)

→ (a^3 – b^3) = (a – b) (a^2 + ab + b^2)

Answer 2

$\sf{\red{\boxed{\bold{Solution}}}}$

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

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$\sf:\implies\:{\bold{ (3p)^3 + \bigg(\dfrac{1}{6}\bigg)^3 + \dfrac{9}{2}p^2 + \dfrac{1}{4}p}}$

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

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

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• From identity

$\sf{\red{\boxed{\bold{a^3 + b^3 + 3ab(a+b) = ( a + b)^3}}}}$

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$\sf:\implies\:{\bold{ ( 3p + \dfrac{1}{6})^3 }}$

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