Suppose x and y are two real numbers with x ≠ y, such that
y = 1+x+x
2, and:
x = 1+y+y
2.
Determine the value of x3+y3.
syms x y
s=solve(@(x,y)y==1+x+x^2 , x==1+y+y^2)
xv=s.x, yv=s.y
xv.^3+yv.^3
eval(ans)
finds that x and y cannot be real as stated
x y
-i -i
i i
- 1 + sqrt(2)*i - 1 - sqrt(2)*i
- 1 - sqrt(2)*i - 1 + sqrt(2)*i
though the last two solutions offer the real sum, 10.
>> quadraticAndCubic
s =
struct with fields:
x: [4×1 sym]
y: [4×1 sym]
xv =
-1i
1i
- 1 + 2^(1/2)*1i
- 1 - 2^(1/2)*1i
yv =
-1i
1i
- 1 - 2^(1/2)*1i
- 1 + 2^(1/2)*1i
ans =
2i
-2i
(- 1 + 2^(1/2)*1i)^3 - (1 + 2^(1/2)*1i)^3
(- 1 + 2^(1/2)*1i)^3 - (1 + 2^(1/2)*1i)^3
ans =
0 + 2i
0 - 2i
10 + 0i
10 + 0i
>>
The x and y pairs solving the equations:
x y
-1i -1i
1i 1i
- 1 + 2^(1/2)*1i - 1 - 2^(1/2)*1i
- 1 - 2^(1/2)*1i - 1 + 2^(1/2)*1i
The sum of the cubes
2i
-2i
(- 1 + 2^(1/2)*1i)^3 - (1 + 2^(1/2)*1i)^3
(- 1 + 2^(1/2)*1i)^3 - (1 + 2^(1/2)*1i)^3
or
2i
-2i
10
10
The sum of the cubes, if real, is 10.
But x and y cannot be real.
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Posted by Charlie
on 2023-10-17 17:50:22 |
computer solution
