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Question Number 49121 by Pk1167156@gmail.com last updated on 03/Dec/18

Answered by mr W last updated on 03/Dec/18

1−((8/h))^3 =(((h−2)/h))^3 1−64((2/h))^3 =(1−(2/h))^3 1−64((2/h))^3 =1−3((2/h))+3((2/h))^2 −((2/h))^3 −3((2/h))+3((2/h))^2 +63((2/h))^3 =0 −3+3((2/h))+63((2/h))^2 =0 ⇒(2/h)=((−3+(√(9+4×3×63)))/(2×63))=(((√(85))−1)/(42)) ⇒h=((84)/((√(85))−1))=(√(85))+1≈10.22 cm

$$\mathrm{1}−\left(\frac{\mathrm{8}}{{h}}\right)^{\mathrm{3}} =\left(\frac{{h}−\mathrm{2}}{{h}}\right)^{\mathrm{3}} \\ $$$$\mathrm{1}−\mathrm{64}\left(\frac{\mathrm{2}}{{h}}\right)^{\mathrm{3}} =\left(\mathrm{1}−\frac{\mathrm{2}}{{h}}\right)^{\mathrm{3}} \\ $$$$\mathrm{1}−\mathrm{64}\left(\frac{\mathrm{2}}{{h}}\right)^{\mathrm{3}} =\mathrm{1}−\mathrm{3}\left(\frac{\mathrm{2}}{{h}}\right)+\mathrm{3}\left(\frac{\mathrm{2}}{{h}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{2}}{{h}}\right)^{\mathrm{3}} \\ $$$$−\mathrm{3}\left(\frac{\mathrm{2}}{{h}}\right)+\mathrm{3}\left(\frac{\mathrm{2}}{{h}}\right)^{\mathrm{2}} +\mathrm{63}\left(\frac{\mathrm{2}}{{h}}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$$−\mathrm{3}+\mathrm{3}\left(\frac{\mathrm{2}}{{h}}\right)+\mathrm{63}\left(\frac{\mathrm{2}}{{h}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{2}}{{h}}=\frac{−\mathrm{3}+\sqrt{\mathrm{9}+\mathrm{4}×\mathrm{3}×\mathrm{63}}}{\mathrm{2}×\mathrm{63}}=\frac{\sqrt{\mathrm{85}}−\mathrm{1}}{\mathrm{42}} \\ $$$$\Rightarrow{h}=\frac{\mathrm{84}}{\sqrt{\mathrm{85}}−\mathrm{1}}=\sqrt{\mathrm{85}}+\mathrm{1}\approx\mathrm{10}.\mathrm{22}\:{cm} \\ $$

Commented by Pk1167156@gmail.com last updated on 06/Dec/18

Thank you sir

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