Ramanujan can do this instantly

Anybody can do this instantly

With a little experimentation or knowledge of the cubes, you will notice that $126=125+1=5^3+1^3$ . Hence, $A=5$ , and $B=1$ (or the other way around, it doesn't matter). $(A+B)^3=6^3=\boxed{216}$ .

The cube nearest to 126 is $5^3=125$ . The difference i.e. $126-125=1$ is also a cube, $1^3=1$ .Therefore, the two numbers are 1 and 5, and their sum is $\underbrace{5}_A+\underbrace{1}_B=6$

Here, let us examine and split 126 into two parts.

We can see that $126=125+1$

Now, we can represent $125=5^{3}$ and $1=1^{3}$

So, $126=125+1=5^{3}+1^{3}$ . So, now 126 is in $A^{3}+B^{3}$ form with A=5 and B=1.

So, $A+B=5+1=\boxed{6}$

the given number is 126 solu: we write 126=5^3+1^3

126=125+1

126=126

$5^3+1^3=126$ . Hence $a+b=6.$

We're looking for two numbers which, when cubed, make 126. As 126 is relatively small in terms of products of cubed numbers, we can just list the cubes of the first few natural numbers and try and spot which two we want:

$1^3 = 1$ $2^3 = 8$ $3^3 = 27$ $4^3 = 64$ $5^3 = 125$

Once we've reached this stage we can tell that $1^3 + 5^3 = 1 + 125 = 126$ So it's fairly obvious that $1 + 5 = \boxed{6}$ is our answer!

(5^3)+(1^3)=126

HELLOW TO ALL . 126 CAN BE EXPRESSED AS
126 = $5^{3}$ + $1^{3}$ . WHERE A = 5 AND B = 1 SO A + B = 6 . SO THE ANSWER IS $\boxed{6}$ .

$126=125+1=5^3+1^3$

$\rightarrow A+B=1+5=\boxed {6}$

There are 5 cubes lesser than 126: 1,8,27,64,125. Taking a^{3}=125, a=5. 126-125=1. 1 is its own cube,so b^{3}=1 and b=1. Therefore, a+b=5+1=6.

$126=125+1=5^3+1^3$

$(5+1)^3=6^3=\fbox{216}$

Seriously....this is like level 0, not even level 1.

We can write 126 as 5^3+1^3 =>5+1=6

(5)3 + (1)3 =125+1=126

Since A^3 + B^3 = 126, Sum of cube of last digit of A and B should be equal to 6. So the pair of last digits could be 1 and 5 or 3 and 9 or 4 and 8. Now pick the first last digit combination 1 and 5. 1^3. + 5^3 itself is 126. Hence A+B=6

public class numbers{ public static void main(String[] args){

  for(int number1 = 0;number1 < 126;number1++){
        for(int number2 = 0;number2 < 126;number2++){
              boolean CUBE = Math.pow(number1, 3) + Math.pow(number2, 3) == 126;

            if(CUBE){
                  System.out.println(number1 + "^3 + " + number2 + "^3 = 126");
            }
         }
     }
 }

}

5 and 1 are the numbers so 5+1=6

5^3+1^3=126= 5+1=6

5^(3)+1^(3) = 125+1 = 126

5^3+1^3=125+1=126; So, A=5 and B=1; That is why, A+B=5+1=6

it is very easy ifu start from 10 and cube them you wil get like this 10cube=1000, 9=729,8=512,7=343,6=216,5=125,4=64,3=27,2=8and1=1 now from which cubing add you get 126 it is 125+1 that is 5+1=6

1^3+5^3

The number 126 can be written as a sum of 125 and 1.
125 can be expressed as A³ and 1 can be expressed as B³.
Therefore to find A and B,
we have to find the cube roots(³√) of 125 and 1.
The ³√125 = 5 and the ³√1 = 1.
Hence, A³ can be written as 5³ and
B³ can be written as 1³.
Therefore, A = 5 and B = 1.
The sum of A and B = 5 + 1 = 6.

starting from 2 keep going then when you reach at 5 you see that the cube of 5 comes 125 and then your mind just clicks that the cube of 1 is always 1 and adding 125 and 1 comes 126 so here is the answer 5 and 1 then adding 1 and 5 comes 6 so 6 is the right answer

# for (int x = 1; x <= 126; x++) for (int y = 126; y >= 1; y--) if ((x x x) + (y y y) == 126) { cout << x << " - " << y << "\n"; }