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Question Number 217643 by Rojarani last updated on 17/Mar/25

a^(1/3) +b^(1/3) +c^(1/3) =(1/(3^(1/3) −2^(1/3) )) b+c−a=?

$$\:{a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} +{c}^{\frac{\mathrm{1}}{\mathrm{3}}} =\frac{\mathrm{1}}{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} } \\ $$$$\:\:{b}+{c}−{a}=? \\ $$

Answered by mehdee7396 last updated on 17/Mar/25

(1/( (3)^(1/3) −(2)^(1/3) ))=(9)^(1/3) +(6)^(1/3) +(4)^(1/3) ⇒ a=9 &b=6&c=4⇒b+c−a=1 or a=6 &b=9&c=4u⇒b+c−a=7 or a=4 &b=9&c=6⇒b+c−a=11 1 or 7 or 11

$$\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}}−\sqrt[{\mathrm{3}}]{\mathrm{2}}}=\sqrt[{\mathrm{3}}]{\mathrm{9}}+\sqrt[{\mathrm{3}}]{\mathrm{6}}+\sqrt[{\mathrm{3}}]{\mathrm{4}}\Rightarrow \\ $$$${a}=\mathrm{9}\:\&{b}=\mathrm{6\&}{c}=\mathrm{4}\Rightarrow{b}+{c}−{a}=\mathrm{1} \\ $$$${or}\:\:{a}=\mathrm{6}\:\&{b}=\mathrm{9\&}{c}=\mathrm{4}{u}\Rightarrow{b}+{c}−{a}=\mathrm{7} \\ $$$${or}\:\:{a}=\mathrm{4}\:\&{b}=\mathrm{9\&}{c}=\mathrm{6}\Rightarrow{b}+{c}−{a}=\mathrm{11} \\ $$$$\mathrm{1}\:{or}\:\:\mathrm{7}\:\:{or}\:\:\mathrm{11} \\ $$

Commented by Rojarani last updated on 17/Mar/25

Sir thanks.

$${Sir}\:{thanks}. \\ $$

Answered by Rasheed.Sindhi last updated on 17/Mar/25

a^(1/3) +b^(1/3) +c^(1/3) =(1/(3^(1/3) −2^(1/3) )) b+c−a=? a^(1/3) +b^(1/3) +c^(1/3) =(1/(3^(1/3) −2^(1/3) )) ∙ ((3^(2/3) +3^(1/3) ∙2^(1/3) +2^(2/3) )/(3^(2/3) +3^(1/3) ∙2^(1/3) +2^(2/3) )) =((9^(1/3) +6^(1/3) +4^(1/3) )/((3^(1/3) )^3 −(2^(1/3) )^3 ))=((9^(1/3) +6^(1/3) +4^(1/3) )/(3−2)) =9^(1/3) +6^(1/3) +4^(1/3) Case a=9 b+c−a=6+4−9=1 ✓ Case a=6 b+c−a=9+4−6=7 ✓ Case a=4 b+c−a=9+6−4=11 ✓ determinant ((((A−B)(A^2 +AB+B^2 )=A^3 −B^3 ^(Formula used) )))

$$\:{a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} +{c}^{\frac{\mathrm{1}}{\mathrm{3}}} =\frac{\mathrm{1}}{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} } \\ $$$$\:\:{b}+{c}−{a}=? \\ $$$$\:{a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} +{c}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} }\:\centerdot\:\frac{\mathrm{3}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \centerdot\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{3}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \centerdot\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} } \\ $$$$=\frac{\mathrm{9}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{6}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\left(\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} −\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} }=\frac{\mathrm{9}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{6}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\mathrm{3}−\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{9}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{6}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${Case}\:{a}=\mathrm{9} \\ $$$${b}+{c}−{a}=\mathrm{6}+\mathrm{4}−\mathrm{9}=\mathrm{1}\:\checkmark \\ $$$${Case}\:{a}=\mathrm{6} \\ $$$${b}+{c}−{a}=\mathrm{9}+\mathrm{4}−\mathrm{6}=\mathrm{7}\:\checkmark \\ $$$${Case}\:{a}=\mathrm{4} \\ $$$${b}+{c}−{a}=\mathrm{9}+\mathrm{6}−\mathrm{4}=\mathrm{11}\:\checkmark \\ $$$$\begin{array}{|c|}{\overset{{Formula}\:{used}} {\left({A}−{B}\right)\left({A}^{\mathrm{2}} +{AB}+{B}^{\mathrm{2}} \right)={A}^{\mathrm{3}} −{B}^{\mathrm{3}} }}\\\hline\end{array} \\ $$

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