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

$$Evaluate$$$${(26+15\sqrt{3})}^{1/3}+{(26-15\sqrt{3})}^{1/3}$$

Answered by deleteduser1 last updated on 29/Jan/24

$$Leta=\sqrt[3]{26+15\sqrt{3}};b=\sqrt[3]{26-15\sqrt{3}}\Rightarrow ab=1$$$$a^3+b^3=52={(a+b)}^3-3ab(a+b)=(a+b)[{(a+b)}^2-3]$$$$x^3-3x-52=0\Rightarrow x=4=a+b=\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}$$

Answered by esmaeil last updated on 29/Jan/24

$$=\mathrm{\Omega }$$$$way2$$$$(26+15\sqrt{3})={(2+\sqrt{3})}^3$$$$(26-15\sqrt{3})={(2-\sqrt{3})}^3\to$$$$\mathrm{\Omega }=4$$

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

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

Commented by necx122 last updated on 30/Jan/24

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

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