find-the-cubic-of-26-15-3-1-3-26-15-3-1-3-

Question Number 172022 by Mikenice last updated on 23/Jun/22

$${find}\:{the}\:{cubic}\:{of} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}}\:\:\:+\:\sqrt[{\mathrm{3}}]{\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}} \\ $$

Answered by Rasheed.Sindhi last updated on 23/Jun/22

Let ((26+15(√3)))^(1/3) + ((26−15(√3)))^(1/3) =a where a∈R (((26+15(√3)))^(1/3) + ((26−15(√3)))^(1/3) )^3 =a^3 determinant ((((a+b)^3 =a^3 +b^3 +3ab(a+b)))) 26+15(√3) +26−15(√3) +3(26+15(√3))(26−15(√3))(a)=a^3 (26+15(√3))(26−15(√3)) =26^2 −3(15^2 )=676−675=1 52+3(1)(a)=a^3 a^3 −3a−52=0 (a−4)(a^2 +4a+13)=0 a=4 ∣ a=((−4±(√(16−52)))/2)=((−4±6i)/2)=−2±3i∉R(Rejected) a=4 (only solution)

$${Let}\:\sqrt[{\mathrm{3}}]{\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}}\:\:\:+\:\sqrt[{\mathrm{3}}]{\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}}\:={a} \\ $$$$\:{where}\:{a}\in\mathbb{R} \\ $$$$\left(\sqrt[{\mathrm{3}}]{\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}}\:\:\:+\:\sqrt[{\mathrm{3}}]{\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}}\:\right)^{\mathrm{3}} ={a}^{\mathrm{3}} \\ $$$$\begin{array}{|c|}{\left({a}+{b}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\mathrm{3}{ab}\left({a}+{b}\right)}\\\hline\end{array} \\ $$$$\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}\:+\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\:+\mathrm{3}\left(\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}\right)\left(\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\right)\left({a}\right)={a}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\left(\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}\right)\left(\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{26}^{\mathrm{2}} −\mathrm{3}\left(\mathrm{15}^{\mathrm{2}} \right)=\mathrm{676}−\mathrm{675}=\mathrm{1} \\ $$$$\mathrm{52}+\mathrm{3}\left(\mathrm{1}\right)\left({a}\right)={a}^{\mathrm{3}} \\ $$$${a}^{\mathrm{3}} −\mathrm{3}{a}−\mathrm{52}=\mathrm{0} \\ $$$$\left({a}−\mathrm{4}\right)\left({a}^{\mathrm{2}} +\mathrm{4}{a}+\mathrm{13}\right)=\mathrm{0} \\ $$$$\:{a}=\mathrm{4}\:\mid\:{a}=\frac{−\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{52}}}{\mathrm{2}}=\frac{−\mathrm{4}\pm\mathrm{6}{i}}{\mathrm{2}}=−\mathrm{2}\pm\mathrm{3}{i}\notin\mathbb{R}\left(\mathcal{R}{ejected}\right) \\ $$$${a}=\mathrm{4}\:\left({only}\:{solution}\right) \\ $$

Commented by Mikenice last updated on 23/Jun/22

$${thanks}\:{sir} \\ $$

Commented by mr W last updated on 23/Jun/22

real+real=real ((26+15(√3)))^(1/3) + ((26−15(√3)))^(1/3) =real ⇒answer can only be real!

$${real}+{real}={real} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}}\:\:\:+\:\sqrt[{\mathrm{3}}]{\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}}\:={real} \\ $$$$\Rightarrow{answer}\:{can}\:{only}\:{be}\:{real}! \\ $$

Commented by Rasheed.Sindhi last updated on 23/Jun/22

$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{sir}}!\:\mathrm{I}'\mathrm{m}\:\mathrm{going}\:\mathrm{to}\:\mathrm{correct}\:\mathrm{now}. \\ $$

Commented by Mikenice last updated on 23/Jun/22