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 174127 by mnjuly1970 last updated on 26/Jul/22

Q: p , q , r are roots of x^( 3) −7x^( 2) =(4−x)(x+2) find the value of : K = (1/( ((qr))^(1/3) )) + (1/( ((pr))^(1/3) )) + (1/( ((pq))^(1/3) )) = ?

$$ \\ $$$$\:\:\:{Q}:\:\:\:\:\:{p}\:,\:{q}\:,\:{r}\:\:\:\:{are}\:\:{roots}\:{of} \\ $$$$\:\:\:\:\:{x}^{\:\mathrm{3}} \:−\mathrm{7}{x}^{\:\mathrm{2}} =\left(\mathrm{4}−{x}\right)\left({x}+\mathrm{2}\right) \\ $$$$\:\:\:\:\:{find}\:\:{the}\:{value}\:{of}\:: \\ $$$$\:\:\:\mathrm{K}\:=\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{qr}}}\:\:+\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{pr}}}\:\:+\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{pq}}}\:=\:? \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 26/Jul/22

x^3 +5x^2 =8x−15−x^2 x^3 +6x^2 −8x+15=0 p+q+r=−6 , pq+qr+rp=−8 , pqr=−15 K = (((p)^(1/3) +(q)^(1/3) +(r)^(1/3) )/( ((pqr))^(1/3) )) ⇒(p)^(1/3) +(q)^(1/3) +(r)^(1/3) = ((pqr))^(1/3) K K^3 = ((((p)^(1/3) +(q)^(1/3) +(r)^(1/3) )/( ((pqr))^(1/3) )))^3 (a+b+c)^3 =a^3 +b^3 +c^3 +3(ab+bc+ca)(a+b+c)−3abc =((p+q+r+3(((pq))^(1/3) +((qr))^(1/3) +((rp))^(1/3) )((p)^(1/3) +(q)^(1/3) +(r)^(1/3) )−3((pqr))^(1/3) )/(pqr)) =((p+q+r+3(((pq))^(1/3) +((qr))^(1/3) +((rp))^(1/3) )(((pqr))^(1/3) K )−3((pqr))^(1/3) )/(pqr)) =((p+q+r+3(((((pqr))^(1/3) )/( (r)^(1/3) ))+((((pqr))^(1/3) )/( (p)^(1/3) )) +((((pqr))^(1/3) )/( (q)^(1/3) )) )(((pqr))^(1/3) K )−3((pqr))^(1/3) )/(pqr)) =((p+q+r+3(((pqr))^(1/3) )^2 (((1 )/( (r)^(1/3) ))+((1 )/( (p)^(1/3) )) +((1 )/( (q)^(1/3) )) )^★ (K )−3((pqr))^(1/3) )/(pqr)) K^3 (pqr)=p+q+r+3(((pqr))^(1/3) )^2 (A )(K )−3((pqr))^(1/3) determinant (((K^3 (pqr)=p+q+r+3(((pqr))^(1/3) )^2 (A )(K )−3((pqr))^(1/3) )))...(I) ^★ A=((1 )/( (r)^(1/3) ))+((1 )/( (p)^(1/3) )) +((1 )/( (q)^(1/3) )) A^3 =(1/r)+(1/p)+(1/q)+3((1/( ((pq))^(1/3) ))+(1/( ((qr))^(1/3) ))+(1/( ((rp))^(1/3) )))(a)−(3/( ((pqr))^(1/3) )) A^3 =(1/r)+(1/p)+(1/q)+3(K)(A)−(3/( ((pqr))^(1/3) )) A^3 =((pq+qr+rp)/(pqr))+3(K)(A)−(3/( ((pqr))^(1/3) )) =((−8)/(−15))+3KA−(3/( ((−15))^(1/3) )) K=((A^3 −(8/(15))+(3/( ((−15))^(1/3) )))/(3A)) determinant (((K=((A^3 −(8/(15))+(3/( ((−15))^(1/3) )))/(3A)))))....(II) Solving (I) & (II) (eqns in boxes) simultaneously we can determine K Too complicated....

$${x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} =\mathrm{8}{x}−\mathrm{15}−{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{15}=\mathrm{0} \\ $$$${p}+{q}+{r}=−\mathrm{6}\:,\:{pq}+{qr}+{rp}=−\mathrm{8}\:,\:{pqr}=−\mathrm{15} \\ $$$$\:\mathrm{K}\:=\:\frac{\sqrt[{\mathrm{3}}]{{p}}\:+\sqrt[{\mathrm{3}}]{{q}}\:+\sqrt[{\mathrm{3}}]{{r}}\:\:}{\:\sqrt[{\mathrm{3}}]{{pqr}}\:}\:\Rightarrow\sqrt[{\mathrm{3}}]{{p}}\:+\sqrt[{\mathrm{3}}]{{q}}\:+\sqrt[{\mathrm{3}}]{{r}}\:=\:\sqrt[{\mathrm{3}}]{{pqr}}\:{K} \\ $$$$ \\ $$$$\:\mathrm{K}^{\mathrm{3}} \:=\:\left(\frac{\sqrt[{\mathrm{3}}]{{p}}\:+\sqrt[{\mathrm{3}}]{{q}}\:+\sqrt[{\mathrm{3}}]{{r}}\:\:}{\:\sqrt[{\mathrm{3}}]{{pqr}}}\right)^{\mathrm{3}} \:\: \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +\mathrm{3}\left({ab}+{bc}+{ca}\right)\left({a}+{b}+{c}\right)−\mathrm{3}{abc} \\ $$$$=\frac{{p}+{q}+{r}+\mathrm{3}\left(\sqrt[{\mathrm{3}}]{{pq}}\:+\sqrt[{\mathrm{3}}]{{qr}}\:+\sqrt[{\mathrm{3}}]{{rp}}\:\right)\left(\sqrt[{\mathrm{3}}]{{p}}\:+\sqrt[{\mathrm{3}}]{{q}}\:+\sqrt[{\mathrm{3}}]{{r}}\:\right)−\mathrm{3}\sqrt[{\mathrm{3}}]{{pqr}}\:}{{pqr}} \\ $$$$=\frac{{p}+{q}+{r}+\mathrm{3}\left(\sqrt[{\mathrm{3}}]{{pq}}\:+\sqrt[{\mathrm{3}}]{{qr}}\:+\sqrt[{\mathrm{3}}]{{rp}}\:\right)\left(\sqrt[{\mathrm{3}}]{{pqr}}\:{K}\:\right)−\mathrm{3}\sqrt[{\mathrm{3}}]{{pqr}}\:}{{pqr}} \\ $$$$=\frac{{p}+{q}+{r}+\mathrm{3}\left(\frac{\sqrt[{\mathrm{3}}]{{pqr}}\:}{\:\sqrt[{\mathrm{3}}]{{r}}}+\frac{\sqrt[{\mathrm{3}}]{{pqr}}\:}{\:\sqrt[{\mathrm{3}}]{{p}}}\:+\frac{\sqrt[{\mathrm{3}}]{{pqr}}\:}{\:\sqrt[{\mathrm{3}}]{{q}}}\:\right)\left(\sqrt[{\mathrm{3}}]{{pqr}}\:{K}\:\right)−\mathrm{3}\sqrt[{\mathrm{3}}]{{pqr}}\:}{{pqr}} \\ $$$$=\frac{{p}+{q}+{r}+\mathrm{3}\left(\sqrt[{\mathrm{3}}]{{pqr}}\:\right)^{\mathrm{2}} \left(\frac{\mathrm{1}\:}{\:\sqrt[{\mathrm{3}}]{{r}}}+\frac{\mathrm{1}\:}{\:\sqrt[{\mathrm{3}}]{{p}}}\:+\frac{\mathrm{1}\:}{\:\sqrt[{\mathrm{3}}]{{q}}}\:\right)^{\bigstar} \left({K}\:\right)−\mathrm{3}\sqrt[{\mathrm{3}}]{{pqr}}\:}{{pqr}} \\ $$$${K}^{\mathrm{3}} \left({pqr}\right)={p}+{q}+{r}+\mathrm{3}\left(\sqrt[{\mathrm{3}}]{{pqr}}\:\right)^{\mathrm{2}} \left({A}\:\right)\left({K}\:\right)−\mathrm{3}\sqrt[{\mathrm{3}}]{{pqr}}\: \\ $$$$\begin{array}{|c|}{{K}^{\mathrm{3}} \left({pqr}\right)={p}+{q}+{r}+\mathrm{3}\left(\sqrt[{\mathrm{3}}]{{pqr}}\:\right)^{\mathrm{2}} \left({A}\:\right)\left({K}\:\right)−\mathrm{3}\sqrt[{\mathrm{3}}]{{pqr}}\:}\\\hline\end{array}...\left(\mathrm{I}\right) \\ $$$$\:^{\bigstar} {A}=\frac{\mathrm{1}\:}{\:\sqrt[{\mathrm{3}}]{{r}}}+\frac{\mathrm{1}\:}{\:\sqrt[{\mathrm{3}}]{{p}}}\:+\frac{\mathrm{1}\:}{\:\sqrt[{\mathrm{3}}]{{q}}} \\ $$$${A}^{\mathrm{3}} =\frac{\mathrm{1}}{{r}}+\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\mathrm{3}\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{pq}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{qr}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{rp}}}\right)\left({a}\right)−\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{{pqr}}} \\ $$$${A}^{\mathrm{3}} =\frac{\mathrm{1}}{{r}}+\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\mathrm{3}\left({K}\right)\left({A}\right)−\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{{pqr}}} \\ $$$${A}^{\mathrm{3}} =\frac{{pq}+{qr}+{rp}}{{pqr}}+\mathrm{3}\left({K}\right)\left({A}\right)−\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{{pqr}}} \\ $$$$\:\:\:=\frac{−\mathrm{8}}{−\mathrm{15}}+\mathrm{3}{KA}−\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{−\mathrm{15}}} \\ $$$${K}=\frac{{A}^{\mathrm{3}} −\frac{\mathrm{8}}{\mathrm{15}}+\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{−\mathrm{15}}}}{\mathrm{3}{A}} \\ $$$$\begin{array}{|c|}{{K}=\frac{{A}^{\mathrm{3}} −\frac{\mathrm{8}}{\mathrm{15}}+\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{−\mathrm{15}}}}{\mathrm{3}{A}}}\\\hline\end{array}....\left(\mathrm{II}\right) \\ $$$${Solving}\:\left(\mathrm{I}\right)\:\&\:\left(\mathrm{II}\right)\:\left({eqns}\:{in}\:{boxes}\right) \\ $$$$\:{simultaneously}\:{we}\:{can}\:{determine}\:{K} \\ $$$$\:\:\:{Too}\:{complicated}.... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com