find the cube root of 9ab^2 + (b^2 +24a^2 )(√(b^2 −3a^2 ))


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Question Number 193221 by York12 last updated on 07/Jun/23

$f i n d t h e c u b e r o o t o f$ $9 {a b}^{2} + (b^{2} + 24 a^{2}) \sqrt{b^{2} - 3 a^{2}}$
	Answered by som(math1967) last updated on 08/Jun/23

	$9 {a b}^{2} + b^{2} \sqrt{b^{2} - 3 a^{2}} + 24 a^{2} \sqrt{b^{2} - 3 a^{2}}$ $9 {a b}^{2} + b^{2} \sqrt{b^{2} - 3 a^{2}} - 3 a^{2} \sqrt{b^{2} - 3 a^{2}}$ $+ 27 a^{2} \sqrt{b^{2} - 3 a^{2}}$ $= (b^{2} - 3 a^{2}) \sqrt{b^{2} - 3 a^{2}} + 3 . {(3 a)}^{2} \sqrt{b^{2} - 3 a^{2}}$ $+ 3.3 a (b^{2} - 3 a^{2}) + 27 a^{3}$ $[∵ 9 {a b}^{2} = 3.3 a (b^{2} - 3 a^{2}) + 27 a^{3}]$ $= {(\sqrt{b^{2} - 3 a^{2}} + 3 a)}^{3}$ $∴ c u b e r o o t = (\sqrt{b^{2} - 3 a^{2}} + 3 a)$
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