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Question
Factorise:
8p3 −\[\frac{27}{p^3}\]
Solution
It is known that,
a3 − b3 = (a − b)(a2 + ab + b2)
\[\ 8p^3 - \frac{27}{p^3}\]
\[ = \left(2p \right)^3 - \left(\frac{3}{p}\right)^3\]
\[ = \left(2p - \frac{3}{p} \right)\left\{\left(2p \right)^2 + \left( \frac{3}{p} \right)^2 + \left(2p \right) \times \left(\frac{3}{p} \right) \right\}\]
\[ = \left(2p - \frac{3}{p} \right)\left(4 p^2 + \frac{9}{p^2} + 6 \right)\]
RELATED QUESTIONS
Simplify:
\[\frac{x^2 - 5x - 24}{\left( x + 3 \right)\left( x + 8 \right)} \times \frac{x^2 - 64}{\left( x - 8 \right)^2}\]
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`16a^3 - 128/b^3`
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Factorise the following:
27x3 – 8y3