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Question Number 141453 by Khalmohmmad last updated on 19/May/21

Commented by PRITHWISH SEN 2 last updated on 19/May/21

$let$ $ab = t a + b = u$ $then$ $tu = 880$ $t + u = 71$ $\Rightarrow t - u = \sqrt{71^{2} - 4 \times 880} = \pm 39$ $\Rightarrow t = 55, u = 16 (cancelling the - ve case$ $a = 11, b = 5 or a = 5, b = 11$ $in either case$ $a^{6} + b^{6} = 1787186$
	Answered by Rasheed.Sindhi last updated on 19/May/21
	$ab(a+b)=880 ∧ ab+a+b=71 ab=71−(a+b) {71−(a+b)}(a+b)=880 (a+b)^2 −71(a+b)+880=0 a+b=((71±(√(71^2 −4(880))))/2) a+b=((71±39)/2)=55,16 ab=71−55,71−16=16,55 (a+b,ab)=(55,16) , (16,55) ^• (a+b,ab)=(55,16) a+b=55 (a+b)^2 =3025 a^2 +b^2 =3025−2ab=3025−2(16)=2993 (a^2 +b^2 )^3 =(2993)^3 a^6 +b^6 +3(ab)^2 (a^2 +b^2 )=2993^3 a^6 +b^6 =2993^3 −3(256)(2993) =26809142033 ^• (a+b,ab)=(16,55) (a+b)^3 =16^3 a^3 +b^3 +3ab(a+b)=16^3 a^3 +b^3 =16^3 −3(55)(16)=1456 (a^3 +b^3 )^2 =1456^2 a^6 +b^6 =1456^2 −2(55)^3 =1787186$
	$a b (a + b) = 880 \land a b + a + b = 71$ $a b = 71 - (a + b)$ ${71 - (a + b)} (a + b) = 880$ ${(a + b)}^{2} - 71 (a + b) + 880 = 0$ $a + b = \frac{71 \pm \sqrt{71^{2} - 4 (880)}}{2}$ $a + b = \frac{71 \pm 39}{2} = 55, 16$ $a b = 71 - 55, 71 - 16 = 16, 55$ $(a + b, a b) = (55, 16), (16, 55)$ $^{∙} (a + b, a b) = (55, 16)$ $a + b = 55$ ${(a + b)}^{2} = 3025$ $a^{2} + b^{2} = 3025 - 2 a b = 3025 - 2 (16) = 2993$ ${(a^{2} + b^{2})}^{3} = {(2993)}^{3}$ $a^{6} + b^{6} + 3 {(a b)}^{2} (a^{2} + b^{2}) = 2993^{3}$ $a^{6} + b^{6} = 2993^{3} - 3 (256) (2993)$ $= 26809142033$ $^{∙} (a + b, a b) = (16, 55)$ ${(a + b)}^{3} = 16^{3}$ $a^{3} + b^{3} + 3 a b (a + b) = 16^{3}$ $a^{3} + b^{3} = 16^{3} - 3 (55) (16) = 1456$ ${(a^{3} + b^{3})}^{2} = 1456^{2}$ $a^{6} + b^{6} = 1456^{2} - 2 {(55)}^{3}$ $= 1787186$
	Answered by Rasheed.Sindhi last updated on 19/May/21

	$a^{6} + b^{6}$ $= (a^{2} + b^{2}) (a^{4} - a^{2} b^{2} + b^{4})$ $= (a^{2} + b^{2}) ({(a^{2} + b^{2})}^{2} - 3 a^{2} b^{2})$ $= ({(a + b)}^{2} - 2 a b) ({({(a + b)}^{2} - 2 a b)}^{2} - 3 {(a b)}^{2})$ $◼ a + b = 55, a b = 16 (i m p o r t e d)$ $= (55^{2} - 2 (16)) ({(55^{2} - 2 (16))}^{2} - 3 {(16)}^{2})$ $= (2993) (2993^{2} - 3 (256))$ $= 26809142033$ $◼ a + b = 16, a b = 55 (i m p o r t e d)$ $a^{6} + b^{6}$ $= ({(a + b)}^{2} - 2 a b) ({({(a + b)}^{2} - 2 a b)}^{2} - 3 {(a b)}^{2})$ $= (16^{2} - 2 (55)) {(16^{2} - 2 (55))}^{2} - 3 {(55)}^{2})$ $= 146 (146^{2} - 3 \times 55^{2})$ $= 1787186$
	Answered by MJS_new last updated on 19/May/21

	$a = x - y \land b = x + y$ $2 x^{3} - 2 {x y}^{2} = 880 \Rightarrow y^{2} = \frac{x^{3} - 440}{x}$ $x^{2} + 2 x - y^{2} = 71 \Rightarrow y^{2} = x^{2} + 2 x - 71$ $\frac{x^{3} - 440}{x} = x^{2} + 2 x - 71$ $x^{2} - \frac{71}{2} x + 220 = 0 \Rightarrow x = 8 \lor x = \frac{55}{2}$ $\Rightarrow y^{2} = 9 \lor y^{2} = \frac{2961}{4}$ $a^{6} + b^{6} = 2 x^{6} + 30 x^{4} y^{2} + 30 x^{2} y^{4} + 2 y^{6}$ $a^{6} + b^{6} = 1787186 \lor 26809142033$
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