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Question Number 150979 by mathdanisur last updated on 17/Aug/21

if x=(4)^(1/3) +(2)^(1/3) +1 find (3/x)+(3/x^2 )+(1/x^3 )=?

$$\mathrm{if}\:\:\mathrm{x}=\sqrt[{\mathrm{3}}]{\mathrm{4}}+\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\:\:\mathrm{find}\:\:\frac{\mathrm{3}}{\mathrm{x}}+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=? \\ $$$$ \\ $$

Answered by john_santu last updated on 18/Aug/21

If x = (4)^(1/3) +(2)^(1/3) +1 then (3/x)+(3/x^2 )+(1/x^3 ) =? a^2 +ab+b^2 = ((a^3 −b^3 )/(a−b)) (2^2 )^(1/3) +(2)^(1/3) .1 +1^2 =((2−1)/( (2)^(1/3) −1))=(1/( (2)^(1/3) −1)) x = (1/( (2)^(1/3) −1)) ; (1/x)=(2)^(1/3) −1 ⇔ (1/x)+1 = (2)^(1/3) ; ((1/x)+1)^3 = 2 ⇔ (1/x^3 ) +(3/x^2 )+(3/x)+1 = 2 ⇔ (1/x^3 )+(3/x^2 )+(3/x) = 1

$$\:\mathrm{If}\:\mathrm{x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{4}}+\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\:\mathrm{then}\:\frac{\mathrm{3}}{\mathrm{x}}+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\:=? \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{ab}+\mathrm{b}^{\mathrm{2}} \:=\:\frac{\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} }{\mathrm{a}−\mathrm{b}}\: \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{2}} }\:+\sqrt[{\mathrm{3}}]{\mathrm{2}}\:.\mathrm{1}\:+\mathrm{1}^{\mathrm{2}} \:=\frac{\mathrm{2}−\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}} \\ $$$$\mathrm{x}\:=\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}}\:;\:\frac{\mathrm{1}}{\mathrm{x}}=\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1} \\ $$$$\Leftrightarrow\:\frac{\mathrm{1}}{\mathrm{x}}+\mathrm{1}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\:;\:\left(\frac{\mathrm{1}}{\mathrm{x}}+\mathrm{1}\right)^{\mathrm{3}} =\:\mathrm{2} \\ $$$$\Leftrightarrow\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\:+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{x}}+\mathrm{1}\:=\:\mathrm{2} \\ $$$$\Leftrightarrow\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{x}}\:=\:\mathrm{1}\: \\ $$

Commented by bramlexs22 last updated on 17/Aug/21

(3/x)=3 (2)^(1/3) −3 ; (3/x^2 ) = 3((2)^(1/3) −1)^2 =3((4)^(1/3) −2(2)^(1/3) +1)= 3 (4)^(1/3) −6(2)^(1/3) +3 (1/x^3 )=((2)^(1/3) −1)^3 = 1−3(4)^(1/3) +3(2)^(1/3) ∴ (3/x)+(3/x^2 )+(1/x^3 ) = 3(2)^(1/3) −3+3(4)^(1/3) −6(2)^(1/3) +3 +1−3(4)^(1/3) +3(2)^(1/3) = −3+3+1 = 1

$$\frac{\mathrm{3}}{\mathrm{x}}=\mathrm{3}\:\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{3}\:; \\ $$$$\:\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{3}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{3}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\mathrm{1}\right)=\:\mathrm{3}\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\:−\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\mathrm{3} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{3}} \:=\:\mathrm{1}−\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{4}}\:+\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{2}}\: \\ $$$$\therefore\:\frac{\mathrm{3}}{\mathrm{x}}+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\:=\:\cancel{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{2}}\:}−\mathrm{3}+\cancel{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{4}}}\:−\cancel{\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{2}}\:}+\mathrm{3}\:+\mathrm{1}−\cancel{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{4}}}\:+\cancel{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{2}}} \\ $$$$\:\:\:\:=\:−\mathrm{3}+\mathrm{3}+\mathrm{1}\:=\:\mathrm{1} \\ $$

Commented by mathdanisur last updated on 17/Aug/21

Thank you Ser

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$

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