Great solution, and nice usage of the identity 3^3 + 4^3 + 5^3 = 6^3. However, the last step kind of irritates in terms of complex analysis, as there are 3 solutions for x. Those solutions are: 606, 606*e^(2i*pi/3) and 606*e^(-2i*pi/3). The general form of the solutions is 606*e^(5ni*pi/6), when n is an integer. Those 3 solutions form a circle when put in the complex plane, which is logical following the e^ix identity. Definitely a topic worth reading about! If anyone has any questions, I really love the subject so will be happy to answer

Solve given Exponential Equation: x³ - 303³ = 404³ + 505³; x = ? x³ = 303³ + 404³ + 505³ = (101³)(3³ + 4³ + 5³) = (101³)(27 + 64 + 125) = (101³)(216) = (101³)(6³) = 606³; x = 606 Answer check: x³ - 303³ = 404³ + 505³; Confirmed as shown Final answer: x = 606

2^a + 2^b + 2^c = 28 = 4(7) = 4(1 + 2 + 4) = 4 + 8 + 16 = 2^2 + 2^3 + 2^4 a = 2 < b = 3 < c = 4

(a + b)^2 = 8ab (a - b)^2 = 4ab [(a + b)/(a - b)]^2 = 2 (a + b)/(a - b) = √2

Nice trick

Math Olympiad: If, 444/(x + 555) = 666; 777/(x + 666) = ? x + 555 = 444/666 = 2/3, x = 2/3 - 555, x + 666 = 2/3 - 555 + 666 = 2/3 + 111 777/(x + 666) = 777/(2/3 + 111) = 3(777)/[2 + 3(111)] = 2331/335

I got all these steps but I was waiting for a trick for the last step. No trick there

Kis class ka hai ye sawal Maine kar liye maths enthusiasts hu . B tech 2nd year

Simpler: You've got a Pythagorean triple. 5^2 - 4^2 = 3^2 So K*5^2 - K*4^2 = K*3^2 Let K= 1111^2 and combine the squared product terms. 5555^2 - 4444^2 = 3333^2

Simpler: You've got a Pythagorean triple. 5^2 - 4^2 = 3^2 So K*5^2 - K*4^2 = K*3^2 Let K= 1111^2 and combine the squared product terms. 5555^2 - 4444^2 = 3333^2 [Edit: Basically what Leo said below.]

😂😂

It's better to write (1111*5)²-(1111*4)² Which will lead you to 1111²(25-16) 1111²(9) Then rewrite 9 as 3² as commonly applied on Pythagorean theorem applications 1111²(3²) And re write the expression 3333² Which is equal to the solution provided in the video, but it comes to show a beautiful expression

Easy 1111^-2 🧐

Excellent 👌

My dumbass would have just brute forced the squares and add them together.

(1111)(9999) = (1111)(10000 - 1) = 11110000 - 1111 = 11108889

You could also simply each k in the expression, eventually reaching k^(7/8)

(9/4)^9/4 (3/2)^9/2 (3/2)^9 × (3/2)^1/2 (19683/512) ×(√3/2) (19683/512) × (√3/2) (6561 ×3 /512) ×(√3/2) (81^2 ×3 /512)×(√3/2) (81^2×3/256×2)×(√3/2) (81^2×3/16^2×2)×(√3/2) (5.0625^2 ×3/2)×(√3/2) im tired ill complete this later

we can also break the exponent into 3/2×3/2...and then do 2 successive breakdowns?

16+8+4 pretty obvious

Nice video 👌

Nice solution 👌

1296 Easy

(√x)^3 = 4^3，√x = 4; x = 16

Much easier if you use substution x=287

Binary me convert krna jyada asaan rahega (28)₁₀ = (11100)₂ = 2⁴+2³+2²

Congratulations! That's awesome! Very good tricks!

√[(1111+1)^2] = 1112

√[(1111+1)^2] = 1112

Nice + helpful

Unnecessary steps after step 2!

if you are going to calculate 513*511, why don't you just calculate 512*512 directly?

513*511 = 250000 + 12000 + 143

Awesome solution 👌

by taking log it would be more simple

Nice one 👍

I am disappointed. The solution was really just a guess and try, and it worked because the solution was so easy. What if, instead of 27, the number was 28? Then the same method would not work. In fact, I found the answer in the same way by assuming the answer was simple and trying the numbers 1, 2, and 3, and 3 worked.

Nice example 👍