Question

$if \: 2x - 4y = 5 \: and \: xy = \frac{1}{3} \: find \: 8 {x}^{3} - 64 {y}^{3}$

Answer 1

Hello,

$2x - 4y = 5 \\ squaring\:\:both\:\:side\:\: ( {2x - 4y)}^{2} = 25 \\ 4 {x}^{2} + 16 {y}^{2} - 16xy = 25 \\ \\ 4 {x}^{2} + 16 {y}^{2} = 25 + 16xy \\ xy = \frac{1}{3} \\ \\ 8 {x}^{3} - 64 {y}^{3}$

Formula of

${a}^{3} - {b}^{3} = (a - b)( {a}^{2} + {b}^{2} + ab) \\ \\ 8 {x}^{3} - 64 {y}^{3} = \\ \\ ( {2x)}^{3} - ( {4y)}^{3} = (2x - 4y)( ({2x})^{2} + ( {4y)}^{2} + (2x)(4y)) \\ \\ = (2x - 4y)( {4x}^{2} + {16y}^{2} + 8xy) \\ = 5(4 {x}^{2} + 16 {y}^{2} + \frac{8}{3} ) \\ \\ = 5(25 + 16xy + \frac{8}{3} ) \\ \\ = 5(25 + \frac{16}{3} + \frac{8}{3} ) \\ \\ = 5( \frac{75 + 16 + 8}{3} ) \\ \\ = 5( \frac{99}{3} ) \\ \\ = 5 \times 33 \\ = 165$
Hope you understand well