then the values of x are:

For an equation built from powers of the same base, substituting u = 3x (with u > 0, since 3x is always positive) turns the exponential equation into an ordinary quadratic in u. Solving that quadratic and converting each valid root back via 3x = u gives the values of x.

1. Rewrite every term to the same base 3: 32x+1 = 3(3x)2, 3x+3 = 27·3x, and 32 = 9.

2. Substitute into the given equation 32x+1 − 3x = 3x+3 − 32 to get 3(3x)2 − 3x = 27·3x − 9.

3. Let u = 3x: 3u2 − u = 27u − 9, which rearranges to 3u2 − 28u + 9 = 0.

4. Solve the quadratic: discriminant = 282 − 4·3·9 = 784 − 108 = 676 = 262, so u = (28 ± 26)/6, giving u = 9 or u = 1/3.

5. Convert back using 3x = u: 3x = 9 = 32 gives x = 2, and 3x = 1/3 = 3−1 gives x = −1.

Both roots satisfy the original equation directly: at x = 2, both sides equal 35 − 32 = 234; at x = −1, both sides equal 3−1 − 3−1 = 0. Since u = 3x must be positive, and both u = 9 and u = 1/3 are positive, neither root is extraneous — so x = 2 and x = −1 are both valid.