Factorize 1-64a3-12a+48a2.

# Factorize $1-64{a}^{3}-12a+48{a}^{2}.$

1. A
${\left(1-4a\right)}^{3}$
2. B
${\left(1+4a\right)}^{3}$
3. C

4. D
${\left(1+2a\right)}^{3}$

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### Solution:

Given, $1-64{a}^{3}-12a+48{a}^{2}.$
We know that the formula,
${\left(x+y\right)}^{3}={x}^{3}+{y}^{3}+3\mathit{xy}\left(x+y\right)$                             or
${\left(x+y\right)}^{3}={x}^{3}+{y}^{3}+3{x}^{2}y+3x{y}^{2}.$ Rewriting the given polynomial,
$1-64{a}^{3}-12a+48{a}^{2}$
${1}^{3}+{\left(-4a\right)}^{3}+3×{1}^{2}×\left(-4a\right)+3×1×{a}^{2}$            …. eq(i)
Comparing eq(i) with the formula, we get,
a=1, b = -4a,
Therefore, eq(i) becomes,
= ${1}^{3}+{\left(-4a\right)}^{3}+3×{1}^{2}×\left(-4a\right)+3×1×{a}^{2}$
= ${\left(1+\left(-4a\right)\right)}^{3}$
= ${\left(1-4a\right)}^{3}$
Therefore, option 1 is correct.

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