Question

If a=1/7-4√3 and b=1/7+4√3, find a³-b³

Answer 1

Heya friend,
Here is the answer you were looking for:
$a = \frac{1}{7 - 4 \sqrt{3} }$

On rationalizing the denominator we get,

$a = \frac{1}{7 - 4 \sqrt{3} } \times \frac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} } \\ \\ a = \frac{7 + 4 \sqrt{3} }{ {(7)}^{2} - {(4 \sqrt{3}) }^{2} } \\ \\ a = \frac{7 + 4 \sqrt{3} }{49 - 48} \\ \\ a = 7 + 4 \sqrt{3}$

$b = \frac{1}{7 + 4 \sqrt{3} }$

On rationalizing the denominator we get,

$b = \frac{1}{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ b = \frac{7 - 4 \sqrt{3} }{ {(7)}^{2} - {(4 \sqrt{3} )}^{2} } \\ \\ b = \frac{7 - 4 \sqrt{3} }{49 - 48} \\ \\ b = 7 - 4 \sqrt{3}$

${a}^{3} - {b}^{3} \\ \\ = {(7 + 4 \sqrt{3} )}^{3} - {(7 - 4 \sqrt{3}) }^{3} \\ \\ = ( {(7)}^{3} + (4 \sqrt{3} )^{3} + 3(7)(4 \sqrt{3} )(7 + 4 \sqrt{3} ) )- ( {(7)}^{3} - {(4 \sqrt{3}) }^{3} - 3(7)(4 \sqrt{3} )(7 - 4 \sqrt{3} )) \\ \\ = (49 + 48 + 84 \sqrt{3} (7 + 4 \sqrt{3} )) - (49 - 48 - 84 \sqrt{3} (7 - 4 \sqrt{3} ) \\ \\ = (49 + 48 + 544 \sqrt{3} + 1008) - (49 - 48 - 544 \sqrt{3} + 1008) \\ \\ = (1105 + 544 \sqrt{3} ) - (1008 - 544 \sqrt{3} \\ \\ = 1105 + 544 \sqrt{3} - 1009 + 544 \sqrt{3} \\ \\ = 96 + 1088 \sqrt{3}$

Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
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