Question Number 57791 by Tawa1 last updated on 12/Apr/19
$$\:\mathrm{If}\:\:\:\:\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:\:=\:\:\mathrm{1}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} \:\:=\:\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{b}^{\mathrm{3}} \:+\:\mathrm{c}^{\mathrm{3}} \:\:=\:\:\mathrm{3}\:\: \\ $$$$\mathrm{then}\:\:\:\:\:\:\mathrm{a}^{\mathrm{5}} \:+\:\mathrm{b}^{\mathrm{5}} \:+\:\mathrm{c}^{\mathrm{5}\:\:} =\:\:? \\ $$
Answered by naka3546 last updated on 12/Apr/19
$$\mathrm{6} \\ $$
Commented by Tawa1 last updated on 12/Apr/19
$$\mathrm{Please}\:\mathrm{workings} \\ $$
Answered by MJS last updated on 12/Apr/19
![det [(a,b,c,1),(a^2 ,b^2 ,c^2 ,2),(a^3 ,b^3 ,c^3 ,3),(a^5 ,b^5 ,c^5 ,x) ]=0 ⇒ x=a^2 bc+ab^2 c+abc^2 −4abc−2a^2 b−2a^2 c−2b^2 c−2ab^2 −2ac^2 −2bc^2 +3a^2 +3b^2 +3c^(2w) +3ab+3ac+3bc (1) c=1−a−b (2) b=((−a+1+(√(−3a^2 +2a+3)))/2) ⇒ x=3a^3 −3a^2 −(3/2)a+((11)/2) (3) 3a^3 −3a^2 −(3/2)a−(1/2)=0 ⇒ x=6](Q57801.png)
$$\mathrm{det}\begin{bmatrix}{{a}}&{{b}}&{{c}}&{\mathrm{1}}\\{{a}^{\mathrm{2}} }&{{b}^{\mathrm{2}} }&{{c}^{\mathrm{2}} }&{\mathrm{2}}\\{{a}^{\mathrm{3}} }&{{b}^{\mathrm{3}} }&{{c}^{\mathrm{3}} }&{\mathrm{3}}\\{{a}^{\mathrm{5}} }&{{b}^{\mathrm{5}} }&{{c}^{\mathrm{5}} }&{{x}}\end{bmatrix}=\mathrm{0} \\ $$$$\Rightarrow\:{x}={a}^{\mathrm{2}} {bc}+{ab}^{\mathrm{2}} {c}+{abc}^{\mathrm{2}} −\mathrm{4}{abc}−\mathrm{2}{a}^{\mathrm{2}} {b}−\mathrm{2}{a}^{\mathrm{2}} {c}−\mathrm{2}{b}^{\mathrm{2}} {c}−\mathrm{2}{ab}^{\mathrm{2}} −\mathrm{2}{ac}^{\mathrm{2}} −\mathrm{2}{bc}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} +\mathrm{3}{c}^{\mathrm{2}{w}} +\mathrm{3}{ab}+\mathrm{3}{ac}+\mathrm{3}{bc} \\ $$$$\left(\mathrm{1}\right)\:{c}=\mathrm{1}−{a}−{b} \\ $$$$\left(\mathrm{2}\right)\:{b}=\frac{−{a}+\mathrm{1}+\sqrt{−\mathrm{3}{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{x}=\mathrm{3}{a}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}{a}+\frac{\mathrm{11}}{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{3}{a}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}{a}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\mathrm{6} \\ $$
Commented by Tawa1 last updated on 12/Apr/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by MJS last updated on 12/Apr/19
$$\mathrm{there}\:\mathrm{should}\:\mathrm{be}\:\mathrm{an}\:\mathrm{easier}\:\mathrm{method}\:\mathrm{but}\:\mathrm{I}\:\mathrm{cannot} \\ $$$$\mathrm{remember}. \\ $$$$\mathrm{anyway}\:\mathrm{this}\:\mathrm{is}\:\mathrm{better}\:\mathrm{than}\:\mathrm{solving}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \\ $$$$\mathrm{degree}\:\mathrm{polynome} \\ $$
Commented by Tawa1 last updated on 12/Apr/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by naka3546 last updated on 12/Apr/19
Commented by Tawa1 last updated on 12/Apr/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{But}\:\mathrm{image}\:\mathrm{too}\:\mathrm{small} \\ $$