Question
Factorise the following: $1-64 a^3-12 a+48 a^2$
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Answer
$1-64 a^3-12 a+48 a^2=(1)^3-(4 a)^3-3 \times 1^2 \times 4 a+3 \times 1 \times(4 a)^2$
${\left[\text { Using identity, }(a-b)^3=a^3-b^3+3 a(-b)(a-b)\right]}$
$=(1-4 a)^3=(1-4 a)(1-4 a)(1-4 a)$
${\left[\text { Using identity, }(a-b)^3=a^3-b^3+3 a(-b)(a-b)\right]}$
$=(1-4 a)^3=(1-4 a)(1-4 a)(1-4 a)$
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