Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 201966 by mnjuly1970 last updated on 17/Dec/23

$$$$$$solve$$$$$$$$\sqrt[3]{x^2+3x+2}(\sqrt[3]{x+1}-\sqrt[3]{x+2})=1$$$$$$

Answered by ajfour last updated on 17/Dec/23

$$letx+\frac{3}{2}=t\Rightarrow p=x+1=t-\frac{1}{2}$$$$q=x+2=t+\frac{1}{2}$$$$\Rightarrow -R=x^2+3x+2=t^2-\frac{1}{4}$$$$pq=-R$$$$(t^2-\frac{1}{4})[-1-\frac{3{\left(pq\right)}^{1/3}}{{(t^2-\frac{1}{4})}^{1/3}}]=1$$$$-1-\frac{3{\left(pq\right)}^{1/3}}{{(t^2-\frac{1}{4})}^{1/3}}=-\frac{1}{R}$$$${(\frac{1}{R}-1)}^3=\frac{27(-R)}{-R}$$$$\Rightarrow \frac{1}{R}-1=3\Rightarrow R=\frac{1}{4}$$$$x^2+3x+2=-\frac{1}{4}$$$$x^2+3x+\frac{9}{4}=0$$$${(x+\frac{3}{2})}^2=0$$$$x=-\frac{3}{2}$$

Commented by mnjuly1970 last updated on 17/Dec/23

$$thanksalotsirajfor$$$$$$

Answered by Frix last updated on 17/Dec/23

$${\left(ab\right)}^{\frac{1}{3}}(a^{\frac{1}{3}}-b^{\frac{1}{3}})=1$$$$\underset{★}{\underbrace{a^{\frac{1}{3}}-b^{\frac{1}{3}}}}={\left(ab\right)}^{-\frac{1}{3}}$$$$a-3{\left(ab\right)}^{\frac{1}{3}}\left(\underset{★}{\underbrace{a^{\frac{1}{3}}-b^{\frac{1}{3}}}}\right)-b={\left(ab\right)}^{-1}$$$$a-b-3=\frac{1}{ab}$$$$a=x+1\wedge b=x+2$$$$x^2+3x+\frac{9}{4}=0$$$$x=-\frac{3}{2}$$$$[\mathrm{Only}\mathrm{true}\mathrm{if}\mathrm{we}\mathrm{stay}\in \mathbb{R}\Rightarrow \sqrt[3]{-r}=-\sqrt[3]{r}]$$

Commented by Frix last updated on 17/Dec/23

$$...\mathrm{and}\mathrm{no}\mathrm{complex}\mathrm{solutions}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com