Shorts... Let x= a^3; y=b^3 then eqn turns to a^6+b^6=65= 2^6+ (-1)^6 And; a+b= 1 Hence ,(a;b)= (2; -1); (-1; 2 ) X; y=(8; -1);( -1; 8) Another questions can be made by Putting a+b= -1 or -3
The given can be a +b = 1 ...(eq1) a^6 + b^6 = 65 ...(eq2) with 3rd-root(x, y) = a, b. Then, squaring (eq1) a^2 +b^2 = 1 -2ab ...(eq3) and cubing (eq3) (a^2 +b^2)^3 = (1 -2ab)^3; a^6 +b^6 +3(a^2+b^2)(ab)^2 = 1 - 6ab +12(ab)^2 -8(ab)^3 and with (eq2) 65 +3(a^2+b^2)(ab)^2 = 1 - 6ab +12(ab)^2 -8(ab)^3. Rearranging with t = ab 2t^3 -9t^2 +6t +64 =0, which can be by SDM, (t +2)(2t^2 -13t +32) = 0 and thus t +2 = 0; t = ab = -2; while 2t^2 -13t +32 = 0 yields t = ab = cmplx. Therefore, a +b = 1; ab = -2 that is a^2 -a -2 = 0; a^2 -2a +a -2 = 0; a(a -2) +(a -2) = 0; (a+1)(a -2) = 0 a = -1, 2 and thus by (eq1) b = 2, -1 That is, (a,b) = (-1, 2), (2, -1) since 3rd-root(x, y) = a, b (x, y) = (-1, 8), (8, -1)
let g1 =a+b-1 (=0), g2=a^6+b^6-65 (=0). Since g1 is 1st order in 'a' and 'b' we can use synthetic division to eliminate one of the components for example using 'a' as reference then remainder g2/g1 = p(b)=2*b^6 - 6*b^5 + 15*b^4 - 20*b^3 + 15*b^2 - 6*b - 64; p(b) = (b - 2)*( b + 1)*( 2*b^4 - 4*b^3 + 15*b^2 - 13*b + 32) so (b=2, a=-1), ( b=-1, a=-2 )
Возведем в куб первое уравнение х+у+3(ху)^(1/3)=1 Второе уравнение (х+у)^2=65+2ху А дальше ху=k^3 И далее как на видео Только на видео к этому моменту пришли на 6 минуте, вместо того чтобы потратить 2 минуты
@@ramunasstulga8264 Just look at the problem. Sum of squares = 65. Sum of cube roots =1. Smooth curves, so clearly there are at most two roots (one positive, one negative). Sum of perfect squares -> text 8 and 1 = 65. Try cube roots -> 2 and 1. Sum must be 1 so +8, and -1. Fits. X and Y are interchangeable. So two solutions. It takes longer to write this than to see the answer.
Shorts...
Let x= a^3; y=b^3 then eqn turns to
a^6+b^6=65= 2^6+ (-1)^6
And; a+b= 1
Hence ,(a;b)= (2; -1); (-1; 2 )
X; y=(8; -1);( -1; 8)
Another questions can be made by Putting a+b= -1 or -3
A wonderful introduction thanks for sharing (x,y)= (8,-1)(-1,8)
The given can be
a +b = 1 ...(eq1)
a^6 + b^6 = 65 ...(eq2)
with 3rd-root(x, y) = a, b.
Then, squaring (eq1)
a^2 +b^2 = 1 -2ab ...(eq3)
and cubing (eq3)
(a^2 +b^2)^3 = (1 -2ab)^3;
a^6 +b^6 +3(a^2+b^2)(ab)^2
= 1 - 6ab +12(ab)^2 -8(ab)^3
and with (eq2)
65 +3(a^2+b^2)(ab)^2
= 1 - 6ab +12(ab)^2 -8(ab)^3.
Rearranging with t = ab
2t^3 -9t^2 +6t +64 =0,
which can be by SDM,
(t +2)(2t^2 -13t +32) = 0
and thus
t +2 = 0; t = ab = -2;
while
2t^2 -13t +32 = 0
yields t = ab = cmplx.
Therefore,
a +b = 1; ab = -2
that is
a^2 -a -2 = 0;
a^2 -2a +a -2 = 0;
a(a -2) +(a -2) = 0;
(a+1)(a -2) = 0
a = -1, 2
and thus by (eq1)
b = 2, -1
That is,
(a,b) = (-1, 2), (2, -1)
since 3rd-root(x, y) = a, b
(x, y) = (-1, 8), (8, -1)
let g1 =a+b-1 (=0), g2=a^6+b^6-65 (=0). Since g1 is 1st order in 'a' and 'b' we can use synthetic division to eliminate one of the components
for example using 'a' as reference then remainder g2/g1 = p(b)=2*b^6 - 6*b^5 + 15*b^4 - 20*b^3 + 15*b^2 - 6*b - 64;
p(b) = (b - 2)*( b + 1)*( 2*b^4 - 4*b^3 + 15*b^2 - 13*b + 32) so (b=2, a=-1), ( b=-1, a=-2 )
Возведем в куб первое уравнение
х+у+3(ху)^(1/3)=1
Второе уравнение
(х+у)^2=65+2ху
А дальше ху=k^3
И далее как на видео
Только на видео к этому моменту пришли на 6 минуте, вместо того чтобы потратить 2 минуты
By observation. x,y = (8,-1), (-1,8)
1/5 marks for correct answer
@@ramunasstulga8264 Just look at the problem. Sum of squares = 65. Sum of cube roots =1. Smooth curves, so clearly there are at most two roots (one positive, one negative). Sum of perfect squares -> text 8 and 1 = 65. Try cube roots -> 2 and 1. Sum must be 1 so +8, and -1. Fits. X and Y are interchangeable. So two solutions. It takes longer to write this than to see the answer.
@@tunneloflight yes, but I mean you would only get one mark for this
@@ramunasstulga8264 Why would I care about marks? Marks are meaningless twaddle.
(X, Y)=( 8; -1); (-1; 8)
(x,y)=(-1,8); (8,-1)
❌ = 8 , -1. 🌱 = -1 , 8
(x,y)=(-1,8),(1,-8),(-8,1),(8,-1)
Math Olympiad Prep: ³√x + ³√y = 1, x² + y² = 65, x, y ϵR; x, y = ?
(³√x + ³√y)³ = (³√x)³ + (³√y)³ + 3(³√xy)(³√x + ³√y) = x + y + 3(³√xy) = 1
(x + y)² = [1 - 3(³√xy)]², x² + y² + 2xy = 65 + 2xy = 1 + 9(³√x²y²) - 6(³√xy)
2xy - 9(³√x²y²) + 6(³√xy) + 64 = 0, Let: u = ³√xy, xy = u³, ³√x²y² = u²; u ϵR
2u³ - 9u² + 6u + 64 = 0, (2u³ + 4u²) - (13u² - 6u - 64) = 0
2u²(u + 2) - (13u - 32)(u + 2) = (u + 2)(2u² - 13u + 32) = 0
2u² - 13u + 32 = 0, Δ = 13² - 8(32) = - 87 < 0, u ϵR; Rejected
u + 2 = 0, u = - 2 = ³√xy; xy = (- 2)³ = - 8, x + y = 1 - 3(³√xy) = 1 + 6 = 7
y = 7 - x, xy = x(7 - x) = 7x - x² = - 8, x² - 7x - 8 = 0, (x + 1)(x - 8) = 0
x + 1 = 0, x = - 1, y = 7 - x = 8 or x - 8 = 0, x = 8, y = 7 - x = - 1
Answer check:
x = - 1, y = 8 or x = 8, y = - 1
³√x + ³√y = - 1 + 2 = 2 - 1 = 1, x² + y² = 1 + 8² = 8² + 1 = 65; Confirmed
Final answer:
x = - 1, y = 8 or x = 8, y = - 1
x⅓ + y⅓ = 1
x² + y² = 65
(x⅓ + y⅓)³ = 1
x + y + 3(xy)⅓(x⅓ + y⅓) = 1
x + y + 3(xy)⅓ = 1
x + y = a
3(xy)⅓ = 1 - a => xy = (1 - a)³/27
x² + y² + 2xy = a²
xy = (a² - 65)/2
(1 - a)³/27 = (a² - 65)/2
2(a - 1)³ + 27(a² - 65) = 0
a = 7 (by inspection)
x + y = 7
xy = (7² - 65)/2 => xy = -8
t² - 7t - 8 = 0
(t + 1)(t - 8) = 0
*(x,y) = { (-1, 8); (8, -1) }*
similar idea
let u = x^(1/3), x=u^3
let v = y^(1/3), y=v^3
u + v = 1
u^6 + v^6 = 65
goal: find u*v
let u*v=a
let f(n) = u^n + v^n
f(0) = 2
f(1) = 1
f(n+1) = f(1)f(n) - u*v*f(n-1) = f(n) - a
f(2) = 1 - 2*a
f(3) = 1 - 3*a
f(4) = 1 - 4*a + 2*a^2
f(5) = 1 - 5*a + 5*a^2
f(6) = 1 - 6*a + 9*a^2 - 2*a^3 = 65
Solve for a: 2*a^3 - 9*a^2 + 6*a -1 = 65
(a + 2)*(2*a^2 - 13*a + 32) = 0
only real solution is a = -2
u*v = -2
u + (-2/u) = 1
u^2 - u - 2 = 0
(u - 2)*(u + 1) = 0
(u,v) = (2,-1), (-1,2)
(x,y) = (8,-1), (-1,8)
(x^2+(y^2)) (1)*(1) =1 (y ➖1x+1) (x)+(y) (33)+(32)=65 (11^11^11) +(11^1110) (11^1^111^1^11^1) (11^1^11^12^5) (1^11^11^1)+(1^11^12^1) = 2^1 (y ➖ 2x+1)
(χ,y) =(8,-1) , (-1,8)