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Question Number 174127 by mnjuly1970 last updated on 26/Jul/22

$$$$$$Q:p,q,rarerootsof$$$$x^3-7x^2=(4-x)(x+2)$$$$findthevalueof:$$$$\mathrm{K}=\frac{1}{\sqrt[3]{qr}}+\frac{1}{\sqrt[3]{pr}}+\frac{1}{\sqrt[3]{pq}}=?$$$$$$

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

$$x^3+5x^2=8x-15-x^2$$$$x^3+6x^2-8x+15=0$$$$p+q+r=-6,pq+qr+rp=-8,pqr=-15$$$$\mathrm{K}=\frac{\sqrt[3]{p}+\sqrt[3]{q}+\sqrt[3]{r}}{\sqrt[3]{pqr}}\Rightarrow \sqrt[3]{p}+\sqrt[3]{q}+\sqrt[3]{r}=\sqrt[3]{pqr}K$$$$$$$${\mathrm{K}}^3={\left(\frac{\sqrt[3]{p}+\sqrt[3]{q}+\sqrt[3]{r}}{\sqrt[3]{pqr}}\right)}^3$$$${(a+b+c)}^3=a^3+b^3+c^3+3(ab+bc+ca)(a+b+c)-3abc$$$$=\frac{p+q+r+3(\sqrt[3]{pq}+\sqrt[3]{qr}+\sqrt[3]{rp})(\sqrt[3]{p}+\sqrt[3]{q}+\sqrt[3]{r})-3\sqrt[3]{pqr}}{pqr}$$$$=\frac{p+q+r+3(\sqrt[3]{pq}+\sqrt[3]{qr}+\sqrt[3]{rp})\left(\sqrt[3]{pqr}K\right)-3\sqrt[3]{pqr}}{pqr}$$$$=\frac{p+q+r+3(\frac{\sqrt[3]{pqr}}{\sqrt[3]{r}}+\frac{\sqrt[3]{pqr}}{\sqrt[3]{p}}+\frac{\sqrt[3]{pqr}}{\sqrt[3]{q}})\left(\sqrt[3]{pqr}K\right)-3\sqrt[3]{pqr}}{pqr}$$$$=\frac{p+q+r+3{\left(\sqrt[3]{pqr}\right)}^2{(\frac{1}{\sqrt[3]{r}}+\frac{1}{\sqrt[3]{p}}+\frac{1}{\sqrt[3]{q}})}^{★}\left(K\right)-3\sqrt[3]{pqr}}{pqr}$$$$K^3\left(pqr\right)=p+q+r+3{\left(\sqrt[3]{pqr}\right)}^2\left(A\right)\left(K\right)-3\sqrt[3]{pqr}$$$$\overline{)\begin{array}{c}{K}^3\left(pqr\right)=p+q+r+3{\left(\sqrt[3]{pqr}\right)}^2\left(A\right)\left(K\right)-3\sqrt[3]{pqr}\end{array}}...\left(\mathrm{I}\right)$$$${}^{★}A=\frac{1}{\sqrt[3]{r}}+\frac{1}{\sqrt[3]{p}}+\frac{1}{\sqrt[3]{q}}$$$$A^3=\frac{1}{r}+\frac{1}{p}+\frac{1}{q}+3(\frac{1}{\sqrt[3]{pq}}+\frac{1}{\sqrt[3]{qr}}+\frac{1}{\sqrt[3]{rp}})\left(a\right)-\frac{3}{\sqrt[3]{pqr}}$$$$A^3=\frac{1}{r}+\frac{1}{p}+\frac{1}{q}+3\left(K\right)\left(A\right)-\frac{3}{\sqrt[3]{pqr}}$$$$A^3=\frac{pq+qr+rp}{pqr}+3\left(K\right)\left(A\right)-\frac{3}{\sqrt[3]{pqr}}$$$$=\frac{-8}{-15}+3KA-\frac{3}{\sqrt[3]{-15}}$$$$K=\frac{A^3-\frac{8}{15}+\frac{3}{\sqrt[3]{-15}}}{3A}$$$$\overline{)\begin{array}{c}K=\frac{A^3-\frac{8}{15}+\frac{3}{\sqrt[3]{-15}}}{3A}\end{array}}....\left(\mathrm{II}\right)$$$$Solving\left(\mathrm{I}\right)\mathrm{\&}\left(\mathrm{II}\right)\left(eqnsinboxes\right)$$$$simultaneouslywecandetermineK$$$$Toocomplicated....$$

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