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Question Number 71903 by naka3546 last updated on 21/Oct/19
ax + by = 3 ax^2 + by^2 = 7 ax^3 + by^3 = 16 ax^4 + by^4 = 42 ax^5 + by^5 = ?
$${ax}\:+\:{by}\:\:=\:\:\mathrm{3} \\ $$$${ax}^{\mathrm{2}} \:+\:{by}^{\mathrm{2}} \:\:=\:\:\mathrm{7} \\ $$$${ax}^{\mathrm{3}} \:+\:{by}^{\mathrm{3}} \:\:=\:\:\mathrm{16} \\ $$$${ax}^{\mathrm{4}} \:+\:{by}^{\mathrm{4}} \:\:=\:\:\mathrm{42} \\ $$$${ax}^{\mathrm{5}} \:+\:{by}^{\mathrm{5}} \:\:=\:\:? \\ $$$$ \\ $$
Answered by mind is power last updated on 21/Oct/19
(x+y)(ax^k +by^k )=ax^(k+1) +by^(k+1) +xy(ax^(k−1) +by^(k−1) ) ⇒ { (((x+y).3=7+xy(a+b)..1)),((7(x+y)=16+xy(3)..2)) :} 16(x+y)=42+xy.(7)...3 2&3 s=x+y and p=xy⇒ 7s=16+3p 16s=42+7p 16(((16+3p)/7))=42+7p ⇒256+48p=294+49p p=−38=xy s=((16−114)/7)=((−98)/7)=−14=x+y (x+y).(ax^4 +by^4 )=ax^5 +by^5 +xy(ax^3 +by^3 ) ⇒−14(42)=ax^5 +by^5 −38(16) ⇒ax^5 +by^5 =38.16−14.42=20
$$\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{ax}^{\mathrm{k}} +\mathrm{by}^{\mathrm{k}} \right)=\mathrm{ax}^{\mathrm{k}+\mathrm{1}} +\mathrm{by}^{\mathrm{k}+\mathrm{1}} +\mathrm{xy}\left(\mathrm{ax}^{\mathrm{k}−\mathrm{1}} +\mathrm{by}^{\mathrm{k}−\mathrm{1}} \right) \\ $$$$\Rightarrow\begin{cases}{\left(\mathrm{x}+\mathrm{y}\right).\mathrm{3}=\mathrm{7}+\mathrm{xy}\left(\mathrm{a}+\mathrm{b}\right)..\mathrm{1}}\\{\mathrm{7}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{16}+\mathrm{xy}\left(\mathrm{3}\right)..\mathrm{2}}\end{cases} \\ $$$$\mathrm{16}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{42}+\mathrm{xy}.\left(\mathrm{7}\right)…\mathrm{3} \\ $$$$\mathrm{2\&3}\:\:\mathrm{s}=\mathrm{x}+\mathrm{y}\:\mathrm{and}\:\mathrm{p}=\mathrm{xy}\Rightarrow \\ $$$$\mathrm{7s}=\mathrm{16}+\mathrm{3p} \\ $$$$\mathrm{16s}=\mathrm{42}+\mathrm{7p} \\ $$$$\mathrm{16}\left(\frac{\mathrm{16}+\mathrm{3p}}{\mathrm{7}}\right)=\mathrm{42}+\mathrm{7p} \\ $$$$\Rightarrow\mathrm{256}+\mathrm{48p}=\mathrm{294}+\mathrm{49p} \\ $$$$\mathrm{p}=−\mathrm{38}=\mathrm{xy} \\ $$$$\mathrm{s}=\frac{\mathrm{16}−\mathrm{114}}{\mathrm{7}}=\frac{−\mathrm{98}}{\mathrm{7}}=−\mathrm{14}=\mathrm{x}+\mathrm{y} \\ $$$$\left(\mathrm{x}+\mathrm{y}\right).\left(\mathrm{ax}^{\mathrm{4}} +\mathrm{by}^{\mathrm{4}} \right)=\mathrm{ax}^{\mathrm{5}} +\mathrm{by}^{\mathrm{5}} +\mathrm{xy}\left(\mathrm{ax}^{\mathrm{3}} +\mathrm{by}^{\mathrm{3}} \right) \\ $$$$\Rightarrow−\mathrm{14}\left(\mathrm{42}\right)=\mathrm{ax}^{\mathrm{5}} +\mathrm{by}^{\mathrm{5}} −\mathrm{38}\left(\mathrm{16}\right) \\ $$$$\Rightarrow\mathrm{ax}^{\mathrm{5}} +\mathrm{by}^{\mathrm{5}} =\mathrm{38}.\mathrm{16}−\mathrm{14}.\mathrm{42}=\mathrm{20} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by MJS last updated on 22/Oct/19
great!
$$\mathrm{great}! \\ $$
Commented by mind is power last updated on 23/Oct/19
thanx sir
$$\mathrm{thanx}\:\mathrm{sir}\: \\ $$
Answered by MJS last updated on 21/Oct/19
solve (1) for y, (2) for b, (3) for a; these are all linear, then (4) for x (quadratic) ⇒ x=−7±(√(87)) y=−7∓(√(87)) a=((49)/(76))±((457(√(87)))/(6612)) b=((49)/(76))∓((457(√(87)))/(6612)) ax^5 +by^5 =20
$$\mathrm{solve}\:\left(\mathrm{1}\right)\:\mathrm{for}\:{y},\:\left(\mathrm{2}\right)\:\mathrm{for}\:{b},\:\left(\mathrm{3}\right)\:\mathrm{for}\:{a};\:\mathrm{these}\:\mathrm{are} \\ $$$$\mathrm{all}\:\mathrm{linear},\:\mathrm{then}\:\left(\mathrm{4}\right)\:\mathrm{for}\:{x}\:\left(\mathrm{quadratic}\right) \\ $$$$\Rightarrow \\ $$$${x}=−\mathrm{7}\pm\sqrt{\mathrm{87}} \\ $$$${y}=−\mathrm{7}\mp\sqrt{\mathrm{87}} \\ $$$${a}=\frac{\mathrm{49}}{\mathrm{76}}\pm\frac{\mathrm{457}\sqrt{\mathrm{87}}}{\mathrm{6612}} \\ $$$${b}=\frac{\mathrm{49}}{\mathrm{76}}\mp\frac{\mathrm{457}\sqrt{\mathrm{87}}}{\mathrm{6612}} \\ $$$${ax}^{\mathrm{5}} +{by}^{\mathrm{5}} =\mathrm{20} \\ $$

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