find-the-cube-root-of-9ab-2-b-2-24a-2-b-2-3a-2-

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)