Question

please ANSWER fast.................

Answer 1

Hey friend, Harish here

Here is your answer:

i) 

$1-64a^3 -12 a + 48a$

⇒ $(1)^3 - (4a)^3 - \bigl [(3)(1)(4a) \left ( 1 - 4a \right ) \bigr ]$

This is of the form (x- y)³ = x³ - y³ - 3xy (x-y) 

Here, x = 1 & y = 4a

Therefore:

⇒ $(1)^3 - (4a)^3 - \bigl [(3)(1)(4a) \left ( 1 - 4a \right ) \bigr ] = (1-4a)^3$

⇒ $(1-4a)(1-4a)(1-4a)$

ii)  

$8p^3 + \frac{12p^2}{3} + \frac{6p}{25} + \frac{1}{125}$

⇒ $8p^3 + \frac{1}{125} + \frac{12p^2}{3} + \frac{6p}{25}$

⇒ $(2p)^3 + \left (\frac{1}{5}\right )^3 +\left [ (3)(2p)\bigl ( \frac{1}{5}\bigr ) \bigl (2p - \frac{1}{5} \bigr ) \right ]$

This of the form (a + b)³: where a = 2p and b = 1/5.

Therefore ;

⇒ $(2p)^3 + \left (\frac{1}{5}\right )^3 +\left [ (3)(2p)\bigl ( \frac{1}{5}\bigr ) \bigl (2p - \frac{1}{5} \bigr ) \right ] = \bigl (2p- \frac{1}{5} \bigr )^3$

⇒ $\bigl (2p- \frac{1}{5} \bigr )\bigl (2p- \frac{1}{5} \bigr )\bigl (2p- \frac{1}{5} \bigr )$
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Hope my answer is helpful to you.