A Nice Math Olympiad Problem | You should know this trick! 

Nice job thanks 🎉🎉🎉

Nice voice! Your videos are always soo nice!

Why not just cube the whole thing?? Using your work, (a-b)³ = 2³ a³ - 3ab(a-b) - b³ = 8 98 - 3*ab*2 = 8 ab = 15 (ab)³ = (x+49)(x-49) = 15³

A really difficult exercise to solve. Congratulations on the solution!

This is the way I solved the problem (I don’t know how to do math notation stuffs, so for this sqrt() is the square root function, crt() is cube root function, a^b means “a to the b power”, - is subtraction, + is addition, / is division and * is multiplication, +pm- means “plus or minus”) Solve for x: crt(x+49)-crt(x-49)=2 Let a = crt(x+49) Let b = crt(x-49) a-b=2 a=2+b a^3 = x + 49 b^3 = x - 49 Isolate variable x and substitute a for b: a^3 + b^3 = x+49+x-49 = 2x (a^3 + b^3)/2 = x (Because a=2+b): ((2+b)^3 + b^3)/2 = x (2+b)^3+b^3 = 2x Finding the polynomial representation of “(2+b)^3”: (2+b)^3 = (2+b)(2+b)(2+b) = (b^2+4b+4)(2+b) = 2b^2+8b+8+b^3+4b^2+4b = b^3+6b^2+12b+8 Creating a quadratic equation: (2+b)^3+b^3 = 2x (2+b)^3 = 2x-b^3 (2+b)^3 = 2x-(x-49) (Because crt(x-49)^3 = x-49) b^3 + 6b^2 + 12b + 8 = 2x - x + 49 b^3 + 6b^2 + 12b + 8 = x+49 b^3 + 6b^2 + 12b + 8 = a^3 6b^2+12b+8 = a^3-b^3 6b^2+12b+8 = x+49 - (x-49) = x-x+49+49 = 98 6b^2+12b+8 = 98 6b^2+12b-90=0 Solving the quadratic equation: 6b^2+12b-90=0 (Quadratic Formula): b = (-12 +pm- sqrt(12^2-4(6)(-90)))/12 b = (-12 +pm- sqrt(144-4(6)(-90)))/12 4*6*-90 = 4*6*-9*10 = 24*-9*10 = -216*10 = 2160 b = (-12 +pm- sqrt(144-(-2160)))/12 b = (-12 +pm- sqrt(2304))/12 b = -1 +pm- sqrt(2304)/12 sqrt(2304) = sqrt(4)*sqrt(576) = sqrt(4)*sqrt(4)*sqrt(144) = 2*2*12 = 48 b= -1 +pm- 48/12 = -1 +pm- 4 b=3 & b=-5 Solving for b=3: (2+3)^3+(3)^3 = 2x (5)^3+(3)^3 = 2x 125 + 27 = 2x 152 = 2x 76 = x Solving for b=-5: (2-5)^3+(-5)^3 =2x (-3)^3 + (-5)^3 = 2x -27 + -125 = 2x -152 = 2x -76 = x x = +pm- 76

You can cube the right and left sides of the equation. After the transformations, we get the equation x^2-49=15^3.

Solved in a similar way but I cubed the equation to get 98-6ab=8 and ab=15.

Didn't know how to solve it until I saw you writing a = (x + 49)^(1/3) and b = (x - 49)^(1/3). Nice idea.

Why not substitute ab=15 into the original equations a=3√(x+49) and b=3√(x-49) where a×b=3√(x^2-49^2). The subsequent equation will be x^2=15^3+49^2=5776, x=+76 or -76.

Nice bro and thanks

-76;76

Х=76

Speaking a language of his own!!! ☹️☹️

X=76

Extremely long and complicated process. I don't think i benefited from it but I'm impressed with your knowledge sir!

Я до сих пор не понимаю какими блудными путями у вас под корнями минус и как вы его достаете оттуда. Оставьте один ответ 76. Или давайте поговорим о мнимой единице и что такое модуль. Ну или, если я не прав, то поправьте кто-нибудь.

台大之所以台大 清蛙之所以清蛙 東蛙之所以東蛙 非常清楚

You lost me after the first substitution.

You could have solved this by inspection rather than going through all this complicated Algebra. This was something my 9th grade Algebra teacher taught us and I have used over the years. Why make your life hard?

Bakbus

By inspection I zeroed in on 5 - 3 = 2, from which ∛125 - √27 = 2. This gives x + 49 = 125 and x - 49 = 27, so x = 76.