64 and 27 are perfect cubes. So, let A = 4^x and B = 3^x Then, numerator = A^3 + B^3 = (A + B) (A^2 -AB + B^2) and denominator = A^2 B + B^2 A = AB (A+B) A+B = 4^x + 3^x > 0 so it can be cancelled to get LHS = (A/B + B/A - 1) = 193/144 = RHS, so A/B + B/A = (193+144)/144 = 377/144 If M^2 = A/B and N^2 = B/A, then (M+N)^2 = A/B + B/A + 2 = 377/144+2 = (377+288)/144 = 625/144 = (25/12)^2, or M + N = +/- 25/12 Similarly, (M - N)^2 = A/B + B/A - 2 = 377/144 - 2 = (377-288)/144 = 49/144 = (7/12)^2, so M - N = +/- 7/12 So, M and N are (25+7)/24 and (25-7)/24 = 4/3 and 3/4 = (4/3 or 3/4)^(x/2), so x/2 = +/- 1 Therefore, x = +/- 2
64 and 27 are perfect cubes. So, let A = 4^x and B = 3^x
Then, numerator = A^3 + B^3 = (A + B) (A^2 -AB + B^2) and denominator = A^2 B + B^2 A = AB (A+B)
A+B = 4^x + 3^x > 0 so it can be cancelled to get LHS = (A/B + B/A - 1) = 193/144 = RHS, so A/B + B/A = (193+144)/144 = 377/144
If M^2 = A/B and N^2 = B/A, then (M+N)^2 = A/B + B/A + 2 = 377/144+2 = (377+288)/144 = 625/144 = (25/12)^2, or M + N = +/- 25/12
Similarly, (M - N)^2 = A/B + B/A - 2 = 377/144 - 2 = (377-288)/144 = 49/144 = (7/12)^2, so M - N = +/- 7/12
So, M and N are (25+7)/24 and (25-7)/24 = 4/3 and 3/4 = (4/3 or 3/4)^(x/2), so x/2 = +/- 1
Therefore, x = +/- 2