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Find real numbers x and y such that x + y = 6 and x^3 + y^3 = 60.

 Sep 26, 2020
 #1
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Find real numbers x and y such that x + y = 6 and x^3 + y^3 = 60.

 

Hello Guest!

 

\(x+y=6\\ y=6-x\\ x^3+y^3=60\\ x^3+(6-x)^3=60\\ x^3+(36-12x+x^2)(6-x)=60\)

\(x^3+216-36x-72x+12x^2+6x^2-x^3=60\\ 18x^2-108x+216=60\)

\(18x^2-108x+156=0\)

\(x = {-108 \pm \sqrt{11664-4\cdot 18\cdot 156} \over 2\cdot 18}\\ x=\frac{-108\pm 20.785}{36}\)

\(x_1=-2.423\\ x_2=-3.577\)      \(y_1=8.423\\ y_2=9.577\)

laugh  !

 Sep 26, 2020
edited by asinus  Sep 27, 2020
 #2
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x + y = 6.......................(1)  

x^3 + y^3 =60.............(2)

y =6 - x

 

Solve for x:
(6 - x)^3 + x^3 = 60

Expand out terms of the left hand side:
18 x^2 - 108 x + 216 = 60

Divide both sides by 18:
x^2 - 6 x + 12 = 10/3

Subtract 12 from both sides:
x^2 - 6 x = -26/3

Add 9 to both sides:
x^2 - 6 x + 9 = 1/3

Write the left hand side as a square:
(x - 3)^2 = 1/3

Take the square root of both sides:
x - 3 = 1/sqrt(3) or x - 3 = -1/sqrt(3)

Add 3 to both sides:
x = 3 + 1/sqrt(3) or x - 3 = -1/sqrt(3)

Add 3 to both sides


x = 3 + 1/sqrt(3)  and y = 3 - 1/sqrt(3)  OR   y = 3 + 1/sqrt(3)  and x = 3 - 1/sqrt(3) 

 Sep 26, 2020

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