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Question Number 119692 by bemath last updated on 26/Oct/20

If x is real number satisfying 3x+(1/(2x))=4 , find the value of 27x^3 +(1/(8x^3 )) .

$${If}\:{x}\:{is}\:{real}\:{number}\:{satisfying} \\ $$$$\mathrm{3}{x}+\frac{\mathrm{1}}{\mathrm{2}{x}}=\mathrm{4}\:,\:{find}\:{the}\:{value}\:{of} \\ $$$$\mathrm{27}{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{8}{x}^{\mathrm{3}} }\:. \\ $$

Commented by bemath last updated on 26/Oct/20

yes gave kudos

$${yes}\:{gave}\:{kudos} \\ $$

Answered by $@y@m last updated on 26/Oct/20

Use formula a^3 +b^3 =(a+b)^3 −3ab(a+b) and try yourself. Ans: 46

$${Use}\:{formula} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\left({a}+{b}\right)^{\mathrm{3}} −\mathrm{3}{ab}\left({a}+{b}\right) \\ $$$${and}\:{try}\:{yourself}. \\ $$$${Ans}:\:\:\mathrm{46} \\ $$

Answered by Ar Brandon last updated on 26/Oct/20

3x+(1/(2x))=4 ⇒ (3x+(1/(2x)))^2 =16 ⇒9x^2 +3+(1/(4x^2 ))=16⇒9x^2 +(1/(4x^2 ))=13 ⇒(9x^2 +(1/(4x^2 )))(3x+(1/(2x)))=13×4 ⇒27x^3 +((9x)/2)+(3/(4x))+(1/(8x^3 ))=52 ⇒(27x^3 +(1/(8x^3 )))=52−(3/2)(3x+(1/(2x)))=52−((3×4)/2) ⇒27x^3 +(1/(8x^3 ))=46

$$\mathrm{3x}+\frac{\mathrm{1}}{\mathrm{2x}}=\mathrm{4}\:\Rightarrow\:\left(\mathrm{3x}+\frac{\mathrm{1}}{\mathrm{2x}}\right)^{\mathrm{2}} =\mathrm{16} \\ $$$$\Rightarrow\mathrm{9x}^{\mathrm{2}} +\mathrm{3}+\frac{\mathrm{1}}{\mathrm{4x}^{\mathrm{2}} }=\mathrm{16}\Rightarrow\mathrm{9x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4x}^{\mathrm{2}} }=\mathrm{13} \\ $$$$\Rightarrow\left(\mathrm{9x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4x}^{\mathrm{2}} }\right)\left(\mathrm{3x}+\frac{\mathrm{1}}{\mathrm{2x}}\right)=\mathrm{13}×\mathrm{4} \\ $$$$\Rightarrow\mathrm{27x}^{\mathrm{3}} +\frac{\mathrm{9x}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4x}}+\frac{\mathrm{1}}{\mathrm{8x}^{\mathrm{3}} }=\mathrm{52} \\ $$$$\Rightarrow\left(\mathrm{27x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{8x}^{\mathrm{3}} }\right)=\mathrm{52}−\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{3x}+\frac{\mathrm{1}}{\mathrm{2x}}\right)=\mathrm{52}−\frac{\mathrm{3}×\mathrm{4}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{27x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{8x}^{\mathrm{3}} }=\mathrm{46} \\ $$

Answered by mathmax by abdo last updated on 26/Oct/20

(3x+(1/(2x)))=4 ⇒(3x+(1/(2x)))^3 =64 ⇒ 27x^3 +3.(3x).(1/(2x))(3x+(1/(2x))) +(1/(8x^3 )) =64 ⇒ 27x^3 +(1/(8x^3 )) +18 =64 ⇒27x^3 +(1/(8x^3 ))=64−18 =46

$$\left(\mathrm{3x}+\frac{\mathrm{1}}{\mathrm{2x}}\right)=\mathrm{4}\:\Rightarrow\left(\mathrm{3x}+\frac{\mathrm{1}}{\mathrm{2x}}\right)^{\mathrm{3}} \:=\mathrm{64}\:\Rightarrow \\ $$$$\mathrm{27x}^{\mathrm{3}} +\mathrm{3}.\left(\mathrm{3x}\right).\frac{\mathrm{1}}{\mathrm{2x}}\left(\mathrm{3x}+\frac{\mathrm{1}}{\mathrm{2x}}\right)\:+\frac{\mathrm{1}}{\mathrm{8x}^{\mathrm{3}} }\:=\mathrm{64}\:\Rightarrow \\ $$$$\mathrm{27x}^{\mathrm{3}} \:+\frac{\mathrm{1}}{\mathrm{8x}^{\mathrm{3}} }\:+\mathrm{18}\:=\mathrm{64}\:\Rightarrow\mathrm{27x}^{\mathrm{3}} \:+\frac{\mathrm{1}}{\mathrm{8x}^{\mathrm{3}} }=\mathrm{64}−\mathrm{18}\:=\mathrm{46} \\ $$

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