Factorize 1 + 64a3 using suitable identity
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1 + 64a^3 , this can also be written in this way also ---
(1)^3 + (4a)^3
[ (a)^3 + (b)^3 = ( a + b)(a^2 + b^2 - ab) ]
here a = 1 and b = 4a
= (1 + 4a)[ (1)^2 + (4a)^2 - (1)(4a) ]
= (1 + 4a)( 1 + 16a^2 - 4a)
(1)^3 + (4a)^3
[ (a)^3 + (b)^3 = ( a + b)(a^2 + b^2 - ab) ]
here a = 1 and b = 4a
= (1 + 4a)[ (1)^2 + (4a)^2 - (1)(4a) ]
= (1 + 4a)( 1 + 16a^2 - 4a)
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