if x^2 = 5x+1 find E = (((x^3 −140) (x^(11) )^(1/(3 )) )/((x^4 +1))^(1/(3 )) )


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Question Number 90212 by john santu last updated on 22/Apr/20

$i f x^{2} = 5 x + 1$ $f i n d E = \frac{(x^{3} - 140) \sqrt[3]{x^{11}}}{\sqrt[3]{x^{4} + 1}}$
Commented by MJS last updated on 22/Apr/20

$\frac{1}{3}$
Commented by jagoll last updated on 22/Apr/20

$how sir ?$
Commented by MJS last updated on 22/Apr/20

$E^{3} = \frac{{(x^{3} - 140)}^{3} x^{11}}{(x^{4} + 1)} = \frac{{(x^{2} x - 140)}^{3} {(x^{2})}^{5} x}{({(x^{2})}^{2} + 1)}$ $x^{2} = 5 x + 1$ $E^{3} = \frac{x {(5 x + 1)}^{5} {(5 x^{2} + x - 140)}^{3}}{25 x^{2} + 10 x + 2}$ $x^{2} - 5 x - 1 = 0 \Rightarrow x = \frac{5 \pm \sqrt{29}}{2}$ $x = \frac{5}{2} + \frac{v}{2}$ $E^{3} = \frac{(v + 5) {(5 v + 27)}^{5} {(5 v^{2} + 52 v - 425)}^{3}}{1024 (25 v^{2} + 270 v + 733)}$ $v^{2} = 29$ $E^{3} = \frac{(v + 5) {(5 v + 27)}^{5} (52 v - 280)}{1024 (270 v + 1458)} =$ $= \frac{(v + 5) {(5 v + 27)}^{5} 4 (13 v - 70)}{1024 \times 54 (5 v + 27)} =$ $= \frac{1}{864} (v + 5) {(5 v + 27)}^{4} {(13 v - 70)}^{3} =$ $= \frac{1}{864} ((v + 5) (13 v - 70)) {({(5 v + 27)}^{2})}^{2} {(13 v - 70)}^{2} =$ $= \frac{1}{864} (13 v^{2} - 5 v - 350) {(25 v^{2} + 270 v + 729)}^{2} (169 v^{2} - 1820 v + 4900) =$ $v^{2} = 29$ $= \frac{1}{864} (27 - 5 v) {(270 v + 1454)}^{2} (9801 - 1820 v) =$ $= \frac{1}{216} (5 v - 27) {(135 v + 727)}^{2} (1820 v - 9801) =$ $= \frac{1}{216} (5 v - 27) (18225 v^{2} + 196290 v + 528529) (1820 v - 9801) =$ $v^{2} = 29$ $\frac{1}{216} (5 v - 27) (196290 v + 1057054) (1820 v - 9801) =$ $= \frac{1}{108} (5 v - 27) (1820 v - 9801) (98145 v + 528527) =$ $= \frac{1}{108} (9100 v^{2} - 98145 v + 264627) (98145 v + 528527) =$ $v^{2} = 29$ $\frac{1}{108} (- 98145 v + 528527) (98145 v + 528527) =$ $= \frac{1}{108} (- 9632441025 v^{2} + 279340789729) =$ $v^{2} = 29$ $= \frac{1}{108} (4) = \frac{1}{27}$ $\Rightarrow E = \frac{1}{3}$
Commented by jagoll last updated on 22/Apr/20

$amazing sir . thank you$
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