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

$$findthecubicof$$$$\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}$$

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

$$Let\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}=a$$$$wherea\in \mathbb{R}$$$${(\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}})}^3=a^3$$$$\overline{)\begin{array}{c}{(a+b)}^3=a^3+b^3+3ab(a+b)\end{array}}$$$$26+15\sqrt{3}+26-15\sqrt{3}+3(26+15\sqrt{3})(26-15\sqrt{3})\left(a\right)=a^3$$$$(26+15\sqrt{3})(26-15\sqrt{3})$$$$={26}^2-3\left({15}^2\right)=676-675=1$$$$52+3\left(1\right)\left(a\right)=a^3$$$$a^3-3a-52=0$$$$(a-4)(a^2+4a+13)=0$$$$a=4\mid a=\frac{-4\pm \sqrt{16-52}}{2}=\frac{-4\pm 6i}{2}=-2\pm 3i\notin \mathbb{R}\left(\mathcal{R}ejected\right)$$$$a=4\left(onlysolution\right)$$

Commented by Mikenice last updated on 23/Jun/22

$$thankssir$$

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

$$real+real=real$$$$\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}=real$$$$\Rightarrow answercanonlybereal!$$

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

$$\mathbb{T}\mathbf{han}\mathbb{k}\mathbf{s}\mathbf{sir}!{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{going}\mathrm{to}\mathrm{correct}\mathrm{now}.$$

Commented by Mikenice last updated on 23/Jun/22

$$butsir[youcanshowyourown$$$$solutiontothatproblem.$$

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

$$i{don}^{\prime }thaveanothersolution.the$$$$solutionfromrasheedisverygood.$$

Commented by Mikenice last updated on 23/Jun/22

$$okaysir$$

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