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Question Number 9874 by j.masanja06@gmail.com last updated on 11/Jan/17

if the eqution x^3 +3hx + q=0 has the roots α,β and γ. find the eqution with the roots of; (i)(1/(αβ)) + (1/(βγ)) + (1/(αγ)) (ii)(1/α^2 ) + (1/β^2 ) + (1/γ^2 )

$$\mathrm{if}\:\mathrm{the}\:\mathrm{eqution}\:\mathrm{x}^{\mathrm{3}} +\mathrm{3hx}\:+\:\mathrm{q}=\mathrm{0}\:\mathrm{has}\:\mathrm{the} \\ $$$$\mathrm{roots}\:\alpha,\beta\:\:\mathrm{and}\:\gamma. \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{eqution}\:\mathrm{with}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}; \\ $$$$\left(\mathrm{i}\right)\frac{\mathrm{1}}{\alpha\beta}\:+\:\frac{\mathrm{1}}{\beta\gamma}\:+\:\frac{\mathrm{1}}{\alpha\gamma} \\ $$$$\left(\mathrm{ii}\right)\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\gamma^{\mathrm{2}} } \\ $$$$ \\ $$

Commented by sandy_suhendra last updated on 11/Jan/17

(i) (1/(αβ)) + (1/(βγ)) + (1/(αγ)) or (1/(αβ)) , (1/(βγ)) and (1/(αγ)) ? (ii) (1/α^2 )+(1/β^2 )+(1/γ_ ^2 ) or (1/α^2 ) , (1/β^2 ) and (1/γ^2 ) ?

$$\left(\mathrm{i}\right)\:\frac{\mathrm{1}}{\alpha\beta}\:+\:\frac{\mathrm{1}}{\beta\gamma}\:+\:\frac{\mathrm{1}}{\alpha\gamma}\:\:\:\:\mathrm{or}\:\:\:\frac{\mathrm{1}}{\alpha\beta}\:\:\:,\:\:\frac{\mathrm{1}}{\beta\gamma}\:\:\mathrm{and}\:\frac{\mathrm{1}}{\alpha\gamma}\:? \\ $$$$\left(\mathrm{ii}\right)\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }+\frac{\mathrm{1}}{\beta^{\mathrm{2}} }+\frac{\mathrm{1}}{\gamma_{} ^{\mathrm{2}} }\:\:\mathrm{or}\:\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:\:,\:\:\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\:\:\mathrm{and}\:\frac{\mathrm{1}}{\gamma^{\mathrm{2}} }\:\:? \\ $$

Commented by j.masanja06@gmail.com last updated on 12/Jan/17

oooh sorry sir has a roots of ; (i)(1/(αβ)) ,(1/(βγ)) and (1/(βγ)) (ii)(1/α^2 ),(1/β^2 ) and (1/γ^2 )

$$\mathrm{oooh}\:\mathrm{sorry}\:\mathrm{sir}\: \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{roots}\:\mathrm{of}\:; \\ $$$$\:\:\:\:\:\:\left(\mathrm{i}\right)\frac{\mathrm{1}}{\alpha\beta}\:,\frac{\mathrm{1}}{\beta\gamma}\:\:\mathrm{and}\:\frac{\mathrm{1}}{\beta\gamma} \\ $$$$\:\:\:\:\:\:\:\:\left(\mathrm{ii}\right)\frac{\mathrm{1}}{\alpha^{\mathrm{2}} },\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\:\mathrm{and}\:\frac{\mathrm{1}}{\gamma^{\mathrm{2}} } \\ $$

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