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प्रश्न
Factorise:
8p3 −\[\frac{27}{p^3}\]
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उत्तर
It is known that,
a3 − b3 = (a − b)(a2 + ab + b2)
\[\ 8p^3 - \frac{27}{p^3}\]
\[ = \left(2p \right)^3 - \left(\frac{3}{p}\right)^3\]
\[ = \left(2p - \frac{3}{p} \right)\left\{\left(2p \right)^2 + \left( \frac{3}{p} \right)^2 + \left(2p \right) \times \left(\frac{3}{p} \right) \right\}\]
\[ = \left(2p - \frac{3}{p} \right)\left(4 p^2 + \frac{9}{p^2} + 6 \right)\]
संबंधित प्रश्न
Simplify:
\[\frac{a^2 + 10a + 21}{a^2 + 6a - 7} \times \frac{a^2 - 1}{a + 3}\]
Simplify:
\[\frac{3 x^2 - x - 2}{x^2 - 7x + 12} \div \frac{3 x^2 - 7x - 6}{x^2 - 4}\]
Simplify:
\[\frac{4 x^2 - 11x + 6}{16 x^2 - 9}\]
Simplify:
\[\frac{a^3 - 27}{5 a^2 - 16a + 3} \div \frac{a^2 + 3a + 9}{25 a^2 - 1}\]
Simplify:
`(1 - 2x + x^2)/(1 - x^3) xx (1 + x + x^2)/(1 + x)`
Factorise:
x3 − 64y3
Factorise:
125y3 − 1
Factorise:
`16a^3 - 128/b^3`
Factorise: 54p3 - 250q3.
Factorise the following:
a6 – 64
