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)

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

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

(X, Y)=( 8; -1); (-1; 8)

❌ = 8 , -1. 🌱 = -1 , 8

(x,y)=(-1,8); (8,-1)

(χ,y) =(8,-1) , (-1,8)

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 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) }*

By observation. x,y = (8,-1), (-1,8)

@ramunasstulga8264

4 күн бұрын

1/5 marks for correct answer

@tunneloflight

4 күн бұрын

@@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.

@ramunasstulga8264

4 күн бұрын

@@tunneloflight yes, but I mean you would only get one mark for this

@tunneloflight

4 күн бұрын

@@ramunasstulga8264 Why would I care about marks? Marks are meaningless twaddle.

(x,y)=(-1,8),(1,-8),(-8,1),(8,-1)

(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)