Does a^3 × b^3 = 3^3 can't be easily solved with fraction 1/3. To be fair. The powers wont be equal value to digit?

a³ x b³ = 3³

(ab)³ = 3³

ab = 3

Put a = 1, b = 3 (or) a = 3, b = 1

(a^3)(b^3)=3^3

=>

(ab)^3=3^3

=>

ab=3

=>

b=3/a

For every given "a" there is one "b"correspond-

ing to it. Thus there are infinitely many (a,b) in R, except a=0 or b=0, satisfying the given equation. e.g. if a=1, then b=3=>(1,3) is a solut-

ion.

(a^3)(b^3) = (ab)^3 = 3^3...(1). (1) holds for ab = 3.

a³×b³ = 3³

(ab)³ = 3³

ab = ∛3³ = 3

Suppose (a,b) = (⅓,9).

a³×b³ = (⅓)³×9³ = (1/27)×729 = 27 = 3³

What are you talking about? The left would just be (ab)^3 using the distributive property. Or, if you leave the problem as it is, if a or b = 1/3:

(1/3)^3 × b^3 = 3^3

3^(-3) × b^3 = 3^3

b^3 = 3^3 / 3^(-3)

b^3 = 3^(3 + 3)

b^3 = 3^6

b = (3^6)^(1/3) = 3^(6 × 1/3) = 3^2 = 9.

As long as ab = 3, 1/3 works fine.

Your question make no sense. Please be articulate and repost.

I can't understand your problem, especially the part about "solved by fraction 1/3".  One possible solution is a = 1, b = 3.  Why don't you like that?

all you have is ( ab)³ = 3³ ===> ab = 3