Here's a much easier solution. Let m and n be x^3 and y^3. Since m+n=35, we have (m+n)^2=m^2+2mn+n^2=1225. Subtracting m^2+n^2=793, we get 2mn=1225-793=432, so mn=216. Then m and n are the roots of the quadratic equation m^2-35m+216=0, which factors as (m-27)(m-8)=0, whose roots are 27 and 8. Finally, x and y are the cube roots of m and n, for which the real values for x and y are 2 and 3 (either order). There are also complex cube roots.
thanks dude. but didn't understood the part from where quadratic equation was formed and also m had values of 27,8 so how you directly put it as equivalent to y , since you haven't calculated value of n and yes I'm bad in maths
This is one of those problems where brute forcing starting with 1 and increasing is faster than the intended method to be used
I appreciate your feedback! Thanks for sharing your perspective. 🙏💕🥰✅
I just looked at the problem for about 10 seconds and knew the answer was 2,3.
x⁶+y⁶=793
x³+y³=35
(x³)²+(y³)²=793
x³+y³=35
Let a=x³ and b=y³
a²+b²=793
a+b=35 -----> b=35-a
a²+(35-a)²=793
a²+a²-70a+1225=793
2a²-70a+432=0
a²-35a+216=0
(a-8)(a-27)=0
Case 1
a-8=0
a=8
8+b=35
b=27
Case 2
a-27=0
a=27
27+b=35
b=8
Case 1: (a, b)=(8, 27)
Recall a=x³ and b=y³
x³=8
x³-8=0
(x-2)(x²+2x+4)=0
x²+2x+4=0
x²+2x+1=-3
(x+1)²=-3
|x+1|=i√3
x+1=±i√3
x=-1±i√3 ❤❤
x-2=0
x=2 ❤
y³=27
y³-27=0
(y-3)(y²+3y+9)=0
y³+3y+9=0
4y³+12y+36=0
4y²+12y+9=-27
(2y+3)²=-27
|2y+3|=3i√3
2y+3=±3i√3
2y=-3±3i√3
y=½(-3±3i√3) ❤❤
y-3=0
y=3 ❤
Case 2: (x, y)=(27, 8)
Recall a=x³ and b=y³
x³=27
x³-27=0
(x-3)(x²+3x+9)=0
x²+3x+9=0
4x²+12x+36=0
4x²+12x+9=-27
(2x+3)²=-27
|2x+3|=3i√3
2x+3=±3i√3
2x=-3±3i√3
x=½(-3±3i√3) ❤❤
x-3=0
x=3 ❤
y³=8
y³-8=0
(y-2)(y²+2y+4)=0
y²+2y+4=0
y²+2y+1=-3
(y+1)²=-3
|y+1|=i√3
y+1=±i√3
y=-1±i√3 ❤❤
y-2=0
y=2 ❤
x³ + y³ = 35 → let: a = x³ and let: b = y³
a + b = 35
x⁶ + y⁶ = 793 → recall: and recall: b = y³
a² + b² = 793
(a + b)² = a² + 2ab + b² → recall: a² + b² = 793
(a + b)² = 2ab + 793 → recall: a + b = 35
35² = 2ab + 793
2ab = 35² - 793
2ab = 432
ab = a = x³ 216 → recall: a + b = 35 → b = 35 - a
a.(35 - a) = 216
35a - a² = 216
a² - 35a + 216 = 0
Δ = (- 35)² - (4 * 216) = 19²
a = (35 ± 19)/2
First case: a = (35 + 19)/2
a = 27 → recall: b = 35 - a
b = 8
Recall: a = x³ → x³ = 27 → x = 3
Recall: b = y³ → y³ = 8 → y = 2
Second case: a = (35 - 19)/2
a = 8 → recall: b = 35 - a
b = 27
Recall: a = x³ → x³ = 8 → x = 2
Recall: b = y³ → y³ = 27→ y = 3
Solution (x ; y)
(3 ; 2)
(2 ; 3)
Thanks for showing your approach ✅💖🙏💕
Here's a much easier solution. Let m and n be x^3 and y^3. Since m+n=35, we have (m+n)^2=m^2+2mn+n^2=1225. Subtracting m^2+n^2=793, we get 2mn=1225-793=432, so mn=216. Then m and n are the roots of the quadratic equation m^2-35m+216=0, which factors as (m-27)(m-8)=0, whose roots are 27 and 8. Finally, x and y are the cube roots of m and n, for which the real values for x and y are 2 and 3 (either order). There are also complex cube roots.
thanks dude. but didn't understood the part from where quadratic equation was formed and also m had values of 27,8 so how you directly put it as equivalent to y , since you haven't calculated value of n
and yes I'm bad in maths
In a quadratic equation x^2+bx+c=0, the sum of the two roots is -b and their product is c.
@@davidellis1929 thanks mate