Question

$\infty answer \: who \: really \: know \: all \\ the \: answers \infty$
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Answer 1

SOLUTION :

$1) \: {x}^{3} + 27 \\ \\ = {x}^{3} + {3}^{3} \\ \\ it \: is \: in \: the \: form \: of \: {a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} ) \\ \\ = > {x}^{3} + {3}^{3} =( x + 3)( {x}^{2} - 3x + 9) \\ \\ 2)8 {x}^{3} + 27 {y}^{3} \\ \\ = (2x) {}^{3} + (3y) {}^{3} \\ \\ = (2x + 3y)(4x {}^{2} - 6xy + 9y {}^{2}) \\ \\ 3) \: \: 343 + 125 {b}^{3} \\ \\ = (7) {}^{3} + (5b) {}^{3} = (7 + 5b)( 49 - 35b + 25 {b}^{2} ) \\ \\ 4) \: \: 1 + 64 {x}^{3} = {(1)}^{3} + (4x) {}^{3} \\ \\ = (1 + 4x)(1 - 4x + 16 {x}^{2} ) \\ \\ 5) \: \: 125 {a}^{3} + \frac{1}{8} \\ \\ = (5 {a})^{3} + ( \frac{1}{2} ) {}^{3} \\ \\ = (5a + \frac{1}{2} )(25 {a}^{2} - \frac{5a}{2} + \frac{1}{4} ) \\ \\ 6) \: \: 216 {x}^{3} + \frac{1}{125} \\ \\ = (6x) {}^{3} + ( \frac{1}{5} ) {}^{3} \\ \\ = (6x + \frac{1}{5} )(36 {x}^{2} - \frac{6x}{5} + \frac{1}{25} )$

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