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Question Number 116221 by bobhans last updated on 02/Oct/20

$$\mathrm{Given}\alpha ,\beta \mathrm{and}\phi \mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}$$$${\mathrm{x}}^3-{\mathrm{px}}^2+\mathrm{qx}-\mathrm{pq}=0.$$$$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+\frac{\beta }{\phi }+\frac{\phi }{\beta }+\frac{\alpha }{\phi }+\frac{\phi }{\alpha }=?$$

Answered by TANMAY PANACEA last updated on 02/Oct/20

$$(\frac{\beta }{\alpha }+\frac{\phi }{\alpha }+1)+(\frac{\alpha }{\beta }+\frac{\phi }{\beta }+1)+(\frac{\alpha }{\phi }+\frac{\beta }{\phi }+1)-3$$$$=(\alpha +\beta +\phi )(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\phi })-3$$$$=(\alpha +\beta +\phi )\left(\frac{\alpha \beta +\beta \phi +\alpha \phi }{\alpha \beta \phi }\right)-3$$$$=p\left(\frac{q}{pq}\right)-3=-2$$

Answered by ruwedkabeh last updated on 02/Oct/20

$$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+\frac{\beta }{\phi }+\frac{\phi }{\beta }+\frac{\alpha }{\phi }+\frac{\phi }{\alpha }$$$$=\frac{{\alpha }^2\phi +{\beta }^2\phi +\alpha {\beta }^2+\alpha {\phi }^2+{\alpha }^2\beta +\beta {\phi }^2}{\alpha \beta \phi }$$$$=\frac{(\alpha +\beta +\phi )(\alpha \beta +\alpha \phi +\beta \phi )-3\alpha \beta \phi }{\alpha \beta \phi }$$$$=\frac{\left(p\right)\left(q\right)-3pq}{pq}$$$$=-2$$$$$$

Answered by bobhans last updated on 02/Oct/20

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