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Question Number 203826 by necx122 last updated on 29/Jan/24

Evaluate (26 + 15(√3))^(1/3) + (26 − 15(√3))^(1/3)

$${Evaluate} \\ $$$$\left(\mathrm{26}\:+\:\mathrm{15}\sqrt{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{3}} +\:\left(\mathrm{26}\:−\:\mathrm{15}\sqrt{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$

Answered by deleteduser1 last updated on 29/Jan/24

Let a=((26+15(√3)))^(1/3) ;b=((26−15(√3)))^(1/3) ⇒ab=1 a^3 +b^3 =52=(a+b)^3 −3ab(a+b)=(a+b)[(a+b)^2 −3] x^3 −3x−52=0⇒x=4=a+b=((26+15(√3)))^(1/3) +((26−15(√3)))^(1/3)

$${Let}\:{a}=\sqrt[{\mathrm{3}}]{\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}};{b}=\sqrt[{\mathrm{3}}]{\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}}\Rightarrow{ab}=\mathrm{1} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{52}=\left({a}+{b}\right)^{\mathrm{3}} −\mathrm{3}{ab}\left({a}+{b}\right)=\left({a}+{b}\right)\left[\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{3}\right] \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}−\mathrm{52}=\mathrm{0}\Rightarrow{x}=\mathrm{4}={a}+{b}=\sqrt[{\mathrm{3}}]{\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}}+\sqrt[{\mathrm{3}}]{\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}} \\ $$

Answered by esmaeil last updated on 29/Jan/24

=Ω way 2 (26+15(√3))=(2+(√3))^3 (26−15(√3))=(2−(√3))^3 → Ω=4

$$=\Omega \\ $$$${way}\:\mathrm{2} \\ $$$$\left(\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}\right)=\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{3}} \\ $$$$\left(\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\right)=\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{3}} \rightarrow \\ $$$$\Omega=\mathrm{4} \\ $$

Answered by Rasheed.Sindhi last updated on 30/Jan/24

(26 + 15(√3)_(a) )^(1/3) + (26 − 15(√3)_(b) )^(1/3) a^(1/3) + b^(1/3) =p (say)....(i) a^(2/3) + b^(2/3) +2a^(1/3) b^(1/3) =p^2 (say) ∵ a^(1/3) b^(1/3) =(ab)^(1/3) =1^(1/3) =1 ∴ a^(2/3) + b^(2/3) =p^2 −2......(ii) (i)×(ii): a+b+a^(1/3) b^(1/3) (a^(1/3) + b^(1/3) )=p^3 −2p a+b+(1)(p)=p^3 −2p p^3 −3p=a+b=26+15(√3) +26−15(√3) =52 p^3 −3p−52=0⇒p=4

$$\left(\underset{{a}} {\underbrace{\mathrm{26}\:+\:\mathrm{15}\sqrt{\mathrm{3}}}}\right)^{\mathrm{1}/\mathrm{3}} +\:\left(\underset{{b}} {\underbrace{\mathrm{26}\:−\:\mathrm{15}\sqrt{\mathrm{3}}}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${a}^{\mathrm{1}/\mathrm{3}} +\:{b}^{\mathrm{1}/\mathrm{3}} ={p}\:\:\left({say}\right)....\left({i}\right) \\ $$$${a}^{\mathrm{2}/\mathrm{3}} +\:{b}^{\mathrm{2}/\mathrm{3}} +\mathrm{2}{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} ={p}^{\mathrm{2}} \:\:\left({say}\right) \\ $$$$\because\:{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} =\left({ab}\right)^{\mathrm{1}/\mathrm{3}} =\mathrm{1}^{\mathrm{1}/\mathrm{3}} =\mathrm{1} \\ $$$$\therefore\:{a}^{\mathrm{2}/\mathrm{3}} +\:{b}^{\mathrm{2}/\mathrm{3}} ={p}^{\mathrm{2}} −\mathrm{2}......\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$${a}+{b}+{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} \left({a}^{\mathrm{1}/\mathrm{3}} +\:{b}^{\mathrm{1}/\mathrm{3}} \right)={p}^{\mathrm{3}} −\mathrm{2}{p} \\ $$$${a}+{b}+\left(\mathrm{1}\right)\left({p}\right)={p}^{\mathrm{3}} −\mathrm{2}{p} \\ $$$${p}^{\mathrm{3}} −\mathrm{3}{p}={a}+{b}=\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}\:+\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\:=\mathrm{52} \\ $$$${p}^{\mathrm{3}} −\mathrm{3}{p}−\mathrm{52}=\mathrm{0}\Rightarrow{p}=\mathrm{4} \\ $$$$\: \\ $$

Commented by necx122 last updated on 30/Jan/24

Thank you so so much sir. This too is wonderful.

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