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Question Number 64542 by LPM last updated on 19/Jul/19

	Answered by MJS last updated on 19/Jul/19

	$b = 1 - a$ $a^{2} + {(1 - a)}^{2} = 2 \Rightarrow a = \frac{1}{2} \pm \frac{\sqrt{3}}{2} \Rightarrow b = \frac{1}{2} \mp \frac{\sqrt{3}}{2}$ $a^{7} = \frac{71}{16} \pm \frac{41 \sqrt{3}}{16} b^{7} = \frac{71}{16} \mp \frac{41 \sqrt{3}}{16}$ $a^{7} + b^{7} = \frac{71}{8}$
	Commented by LPM last updated on 19/Jul/19

	$good$
	Answered by mr W last updated on 19/Jul/19
	$a+b=1 a^2 +b^2 +2ab=1 2+2ab=1 ⇒ab=−(1/2) a^7 +b^7 =(a+b)(a^6 −a^5 b+a^4 b^2 −a^3 b^3 +a^2 b^4 −ab^5 +b^6 ) =a^6 +(1/2)a^4 +(1/4)a^2 +(1/8)+(1/4)b^2 +(1/2)b^4 +b^6 =(a^6 +b^6 )+(1/2)(a^4 +b^4 )+(1/4)(a^2 +b^2 )+(1/8) =(a^3 +b^3 )^2 −2a^3 b^3 +(1/2)[(a^2 +b^2 )^2 −2a^2 b^2 ]+(1/2)+(1/8) =(a^3 +b^3 )^2 +(1/4)+(1/2)[4−(1/2)]+(1/2)+(1/8) =(a^3 +b^3 )^2 +((21)/8) ={(a+b)[(a+b)^2 −3ab]}^2 +((21)/8) ={1+(3/2)}^2 +((21)/8) =((25)/4)+((21)/8) =((71)/8)$
	$a + b = 1$ $a^{2} + b^{2} + 2 a b = 1$ $2 + 2 a b = 1$ $\Rightarrow a b = - \frac{1}{2}$ $a^{7} + b^{7} = (a + b) (a^{6} - a^{5} b + a^{4} b^{2} - a^{3} b^{3} + a^{2} b^{4} - {a b}^{5} + b^{6})$ $= a^{6} + \frac{1}{2} a^{4} + \frac{1}{4} a^{2} + \frac{1}{8} + \frac{1}{4} b^{2} + \frac{1}{2} b^{4} + b^{6}$ $= (a^{6} + b^{6}) + \frac{1}{2} (a^{4} + b^{4}) + \frac{1}{4} (a^{2} + b^{2}) + \frac{1}{8}$ $= {(a^{3} + b^{3})}^{2} - 2 a^{3} b^{3} + \frac{1}{2} [{(a^{2} + b^{2})}^{2} - 2 a^{2} b^{2}] + \frac{1}{2} + \frac{1}{8}$ $= {(a^{3} + b^{3})}^{2} + \frac{1}{4} + \frac{1}{2} [4 - \frac{1}{2}] + \frac{1}{2} + \frac{1}{8}$ $= {(a^{3} + b^{3})}^{2} + \frac{21}{8}$ $= {(a + b) [{(a + b)}^{2} - 3 a b]}^{2} + \frac{21}{8}$ $= {1 + \frac{3}{2}}^{2} + \frac{21}{8}$ $= \frac{25}{4} + \frac{21}{8}$ $= \frac{71}{8}$
	Commented by LPM last updated on 19/Jul/19

	$good$
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