Question Number 51772 by Cheyboy last updated on 30/Dec/18
$$\mathrm{x}^{\mathrm{3}} +\mathrm{12x}+\mathrm{12}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Express}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{surd}\:\mathrm{form} \\ $$$$ \\ $$
Answered by ajfour last updated on 30/Dec/18
$${x}=\left(−\mathrm{6}+\sqrt{\mathrm{36}+\mathrm{64}}\right)^{\mathrm{1}/\mathrm{3}} −\left(\mathrm{6}+\sqrt{\mathrm{36}+\mathrm{64}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:=\:\sqrt[{\mathrm{3}}]{\mathrm{4}}−\sqrt[{\mathrm{3}}]{\mathrm{16}}\:.\: \\ $$
Commented by Cheyboy last updated on 30/Dec/18
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{but}\:\mathrm{i}\:\mathrm{would}\:\mathrm{appreciate} \\ $$$$\mathrm{to}\:\mathrm{know}\:\mathrm{the}\:\mathrm{working} \\ $$
Answered by mr W last updated on 31/Dec/18
![x^3 +12x+12=0 cubic eqn. of this kind can be solved in following way: we know that u^3 +v^3 =(u+v)(u^2 −uv+v^2 )=(u+v)[(u+v)^2 −3uv] ⇒ (u+v)^3 −3uv(u+v)−(u^3 +v^3 )=0 when you compare this with our eqn. you can see that if we put 3uv=−12, u^3 +v^3 =−12, then u+v is the solution of our eqn. 3uv=−12⇒v=−(4/u) u^3 +v^3 =−12⇒u^3 −(4^3 /u^3 )+12=0 ⇒u^6 +12u^3 −64=0 (quadratic eqn. for u^3 ) ⇒(u^3 +16)(u^3 −4)=0 ⇒u^3 = 4 (−16 will also do, but gives the same) ⇒u=(4)^(1/3) ⇒v=−(4/( (4)^(1/3) ))=−2(2)^(1/3) ⇒u+v=(4)^(1/3) −2(2)^(1/3) i.e. the solution of x^3 +12x+12=0 is x=(4)^(1/3) −2(2)^(1/3) ≈−0.9324](https://www.tinkutara.com/question/Q51827.png)
$${x}^{\mathrm{3}} +\mathrm{12}{x}+\mathrm{12}=\mathrm{0} \\ $$$${cubic}\:{eqn}.\:{of}\:{this}\:{kind}\:{can}\:{be}\:{solved} \\ $$$${in}\:{following}\:{way}: \\ $$$$ \\ $$$${we}\:{know}\:{that} \\ $$$${u}^{\mathrm{3}} +{v}^{\mathrm{3}} =\left({u}+{v}\right)\left({u}^{\mathrm{2}} −{uv}+{v}^{\mathrm{2}} \right)=\left({u}+{v}\right)\left[\left({u}+{v}\right)^{\mathrm{2}} −\mathrm{3}{uv}\right] \\ $$$$\Rightarrow\:\left({u}+{v}\right)^{\mathrm{3}} −\mathrm{3}{uv}\left({u}+{v}\right)−\left({u}^{\mathrm{3}} +{v}^{\mathrm{3}} \right)=\mathrm{0} \\ $$$$ \\ $$$${when}\:{you}\:{compare}\:{this}\:{with}\:{our}\:{eqn}. \\ $$$${you}\:{can}\:{see}\:{that}\: \\ $$$${if}\:{we}\:{put}\:\mathrm{3}{uv}=−\mathrm{12},\:{u}^{\mathrm{3}} +{v}^{\mathrm{3}} =−\mathrm{12}, \\ $$$${then}\:{u}+{v}\:{is}\:{the}\:{solution}\:{of}\:{our}\:{eqn}. \\ $$$$ \\ $$$$\mathrm{3}{uv}=−\mathrm{12}\Rightarrow{v}=−\frac{\mathrm{4}}{{u}} \\ $$$${u}^{\mathrm{3}} +{v}^{\mathrm{3}} =−\mathrm{12}\Rightarrow{u}^{\mathrm{3}} −\frac{\mathrm{4}^{\mathrm{3}} }{{u}^{\mathrm{3}} }+\mathrm{12}=\mathrm{0} \\ $$$$\Rightarrow{u}^{\mathrm{6}} +\mathrm{12}{u}^{\mathrm{3}} −\mathrm{64}=\mathrm{0}\:\:\:\left({quadratic}\:{eqn}.\:{for}\:{u}^{\mathrm{3}} \right) \\ $$$$\Rightarrow\left({u}^{\mathrm{3}} +\mathrm{16}\right)\left({u}^{\mathrm{3}} −\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow{u}^{\mathrm{3}} =\:\mathrm{4}\:\:\:\left(−\mathrm{16}\:{will}\:{also}\:{do},\:{but}\:{gives}\:{the}\:{same}\right) \\ $$$$\Rightarrow{u}=\sqrt[{\mathrm{3}}]{\mathrm{4}} \\ $$$$\Rightarrow{v}=−\frac{\mathrm{4}}{\:\sqrt[{\mathrm{3}}]{\mathrm{4}}}=−\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}} \\ $$$$\Rightarrow{u}+{v}=\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}} \\ $$$$ \\ $$$${i}.{e}.\:{the}\:{solution}\:{of}\:{x}^{\mathrm{3}} +\mathrm{12}{x}+\mathrm{12}=\mathrm{0}\:{is} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}}\approx−\mathrm{0}.\mathrm{9324} \\ $$
Commented by Cheyboy last updated on 30/Dec/18
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir} \\ $$$$ \\ $$