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Category: Algebra

proof-that-a-2-a-b-a-c-b-2-b-c-b-a-c-2-c-a-c-b-a-b-c-

Question Number 30456 by daffa123 last updated on 22/Feb/18 $${proof}\:{that} \\ $$$$\frac{{a}^{\mathrm{2}} }{\left({a}−{b}\right)\left({a}−{c}\right)}+\frac{{b}^{\mathrm{2}} }{\left({b}−{c}\right)\left({b}−{a}\right)}+\frac{{c}^{\mathrm{2}} }{\left({c}−{a}\right)\left({c}−{b}\right)}=\:{a}+{b}+{c} \\ $$ Answered by Rasheed.Sindhi last updated on 22/Feb/18 $$\frac{{a}^{\mathrm{2}}…

x-6-6x-5-ax-4-bx-3-cx-2-dx-1-0-all-the-roots-of-the-equation-are-positive-find-a-b-c-d-

Question Number 161500 by HongKing last updated on 18/Dec/21 $$\mathrm{x}^{\mathrm{6}} \:-\:\mathrm{6x}^{\mathrm{5}} \:+\:\mathrm{ax}^{\mathrm{4}} \:+\:\mathrm{bx}^{\mathrm{3}} \:+\:\mathrm{cx}^{\mathrm{2}} \:+\:\mathrm{dx}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{all}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{are}\:\mathrm{positive} \\ $$$$\mathrm{find}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=? \\ $$ Answered by mr W…

1-a-1-b-9-1-a-1-3-1-b-1-3-1-1-a-1-3-1-1-b-1-3-18-8a-4b-

Question Number 161484 by bobhans last updated on 18/Dec/21 $$\:\:\begin{cases}{\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}=\mathrm{9}}\\{\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{b}}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{a}}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{b}}}\right)=\mathrm{18}}\end{cases} \\ $$$$\:\:\:\mathrm{8a}+\mathrm{4b}=? \\ $$ Answered by mr W last updated on 18/Dec/21 $${let}\:{A}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}},\:{B}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}} \\ $$$${let}\:{p}={A}+{B},\:{q}={AB}…