A's and B's

Because $a$ and $b$ are positive integers this means that $a > 0$ and $b > 0$ .

$18a=b^3 \Rightarrow b = \sqrt[3]{18a}$

We know that $18 = 2*3^2$ .

Now we need to find some value of $a$ , so that the number $18*a$ will be of the form $x^3*y^3*z^3...$ where $x,y,z,...$ are all positive integers. We need this form of the number to extract the cubic root as a positive integer since $b$ is a positive integer.

There is an infinite number of this values, but we are interested in the minimum value. This means that $x,y,z,...$ must be minimal positive integers.

To find the minimal value, we look at the factors of number $18$ and see what numbers are missing , so we can extract the integer cubic root. We see that we need the product of numbers $3$ and $4$ to obtain the number $216 = 2^3*3^3$ . Any other product would lead us to a number larger then $216$ and thus, $a$ and $b$ would not have minimum values.

This means that $a$ is the product of the added numbers, $a = 3*4 = 12 (>0)$ and $b$ is the value of the cubic root, $b = \sqrt[3]{216} = 6 (>0)$ .

Because $a$ and $b$ are minimal values, the sum $a + b$ will also be minimal: $a + b = 18$ .

In conclusion, the minimal sum is the number $18$ .

Since $a = {b^3}/{18}$ , we just need to find the minimum $b^3$ which is divisible by $18$ and the corresponding $a$ would be automatically minimum. We look at the decomposition of $18$ that is $2 \cdot 3^2$ . To complete the cube we multiply it by $2^2 \cdot 3 = 12.$ Hence we see, $b^3 = 18 \cdot 12 = 6^3$ and $a = 12$ . Hence $12 + 6 = 18$ is our answer.

Given: 18a=b^3 Checking parity we see that b must be even. Let b=2m for some positive integer m. So we have: 18a=8m^3 i.e. 9a=4m^3 Checking parity we see that a must be even. If a=2k (for some positive integer k) , RHS is still an even term. So we can assume a=4p for some positive integer p. So 9p=m^3 Checking parity , m=3z for some positive integer z. So p=3z^3 We see that minimum solution is available for z=1 & p=3. Now p=3 => a=12 & z=1 => m=3 => b=6 So minimum possible value of a+b is 12+6 i.e. 18

Note that $b^3$ = $(2)*(3^2)*a$ , so $b$ has 2*3=6 as a factor. So $b=6$ is the minimum value of $b$ and thus $a=12$ so that $a+b=18$

18*12=6^3 216=216 :12+6=18

First, find all of the prime factors of 18, which are 2, 3, and 3. Second, find the LCM of 2 and 3, which is 6. So, you have to multiply each prime factor by another number to get 6. So, you multiply 2 into 3 to get 6, and you multiply 3 into 2 to get 6. Since you multiplied 3, 2, and 2 into those numbers, now you have 6^3. So, you multiply 3, 2, and 2 together to get 12, and add 6, to get 18.

AS STATED IN THE QUESTION 18A=B^3 SO WE WILL FIND THE THE SMALLEST NUMBER BY WHICH 18 SHOULD BE DIVIDED SO THAT THE ANSWER IS A PERFECT CUBE WHEN WE DO THE FACTORS OF 18 WE FIND THAT 12 SHOULD BE MULTIPLIED TO MAKE IT A PERFECT CUBE. A=12. AFTER THIS WE CUBE ROOT THIS AND WE FIND 6 AS 18*12=6^3 .A=12 B=6 AFTER ADDING A+B WE WILL FIND THAT 12+6=18

1 8 27 .. .. 216 that is divisible by 18 therefore a=12 as 18x12=216 and b^3=216 therefore b=6

The divisors of 18 is 3,3 and 2. To make it a cube numbers, there must be sets of 3 same divisors. Therefore, multiply 18 with 3, 2 and 2 to make it into a set of 3 3s and a set of 3 2s. 3* 2* 2=12(value of a) 18*12=216 216 is the cube number of 6(value of b) 12+6=18

if we want to get 'b' as a natural no. we have to check the lowest possible number to be multiplied with '18' to get a cube.This can give us lowest possible value.i bet u can solve the problem in 1 minute

18a= $b^3$

b= $sqrt[3]{18a}$

prime factorisation of 18 is 2 * 3 * 3

so we need a 3 and 2 2's to make it a perfect cube.

so a=3 * 2 * 2

b= $sqrt[3]{18a}$

b= $sqrt[3]{216}$

b=6

a + b = 12 + 6

a + b = 18

We have $18 = 3 \times 3 \times 2 = 3 ^2 2$ . Let $a = 3 ^{p_1} 2 ^{p_2} r$ , where $p_1$ and $p_2$ are non-negative integers and $r$ is a positive integer that is coprime to $2$ and $3$ . Since $b^3 = 18 a = 3 ^{2+p_1} 2 ^{1+p_2} r$ is a perfect cube, each of these terms (in the 'prime factorization') must be a perfect cube. Thus $2 + p_1 \equiv 0 \pmod {3}$ , $1 + p_2 \equiv 0 \pmod {3}$ . Therefore $p_1 = 1$ , $p_2 = 2$ and $r = 1$ gives the smallest possible value of $b$ and we have $a = 3 \times 2 \times 2 = 12$ and $b = 3 \times 2 = 6$ . Hence $a + b = 12 + 6 = 18$ .

the least value of b should make a a factor of perfect cube for b and only 12 satisfies to make b a perfect cube which is 216 which is 6

For the least value of $a+b$ ,we must choose value of $a$ in $18a=b^3$ so that $b^3$ has the $\text{least value possible}$ . $18=2\times3^2$ so $a$ must be $2^2\times3=\boxed{12}$ for $b^3$ to have the least possible value. $(216)$ so $a=12\;and\;b=\sqrt[3]{b^3}=\sqrt[3]{216}=\boxed{6}$ so $a+b=12+6=\Large{18}$