Find the value of $a^{3} - 27 b^{3}$ if $a - 3 b = - 6$ and $a b = - 10$

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Solution

$a - 3 b = - 6$
Squaring both sides,
$=> (a - 3 b)^{2} = - 6^{2}$
$=> a^{2} + 9 b^{2} - 2 (a) (3 b) = 36$
$=> a^{2} + 9 b^{2} - 6 (a b) = 36$
$=> a^{2} + 9 b^{2} - 6 (- 10) = 36$
$=> a^{2} + 9 b^{2} = 36 - 60$
$=> a^{2} + 9 b^{2} = - 24$
Now,
$a^{3} - (3 b)^{3} = (a - 3 b) [(a)^{2} + (3 b)^{2} + (a) (3 b)]$
$a^{3} - 27 b^{3} = - 6 (- 24 + 3 (- 10))$
$= - 6 (- 24 - 30)$
$= 324$

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