Okay, so I need to find the greatest integer less than (sqrt(7) + sqrt(5))^6. Hmm, that seems a bit tricky without a calculator. Let me think about how to approach this.

First, maybe I can use the binomial theorem to expand (sqrt(7) + sqrt(5))^6. The binomial theorem says that (a + b)^n = Σ (from k=0 to n) C(n, k) * a^(n-k) * b^k. So here, a is sqrt(7), b is sqrt(5), and n is 6. That should give me all the terms. Let me try expanding that.

But before I dive into expanding everything, maybe there's a smarter way. I remember that expressions like (sqrt(a) + sqrt(b))^n can sometimes be related to their conjugates. The conjugate here would be (sqrt(7) - sqrt(5))^6. If I add those two expressions together, maybe the irrational parts cancel out, leaving an integer. Then, if I can find that sum, subtract 1, and take the floor, that might give me the answer.

Let me test this idea. Let me denote x = sqrt(7) + sqrt(5) and y = sqrt(7) - sqrt(5). Then, x + y = 2*sqrt(7) and x - y = 2*sqrt(5). If I compute x^6 + y^6, maybe that's an integer. Let me check.

But first, let's compute x^6 + y^6. Since y is the conjugate, maybe their powers will add up to an integer. Let's see. To compute x^6 + y^6, perhaps I can find a recursion or a pattern.

Alternatively, I can compute x^6 + y^6 by first computing x^2 and y^2, then building up to higher powers. Let's try that.

First, compute x^2 and y^2:

x^2 = (sqrt(7) + sqrt(5))^2 = 7 + 2*sqrt(35) + 5 = 12 + 2*sqrt(35)

y^2 = (sqrt(7) - sqrt(5))^2 = 7 - 2*sqrt(35) + 5 = 12 - 2*sqrt(35)

So x^2 + y^2 = (12 + 2*sqrt(35)) + (12 - 2*sqrt(35)) = 24

x^2 * y^2 = (12 + 2*sqrt(35))(12 - 2*sqrt(35)) = 144 - (2*sqrt(35))^2 = 144 - 4*35 = 144 - 140 = 4

So (x^2)(y^2) = 4

Now, maybe we can find a recurrence relation for x^n + y^n. Let me denote z_n = x^n + y^n. Then, we can find a recurrence relation for z_n.

We know that z_0 = x^0 + y^0 = 1 + 1 = 2

z_1 = x + y = 2*sqrt(7)

z_2 = x^2 + y^2 = 24

z_3 = x^3 + y^3 = (x + y)(x^2 + y^2 - xy) + xy(x + y). Wait, maybe a better way is to use the recurrence relation.

Since x and y are roots of the equation t^2 - (x + y)t + xy = 0. But x + y = 2*sqrt(7), and xy = (sqrt(7) + sqrt(5))(sqrt(7) - sqrt(5)) = 7 - 5 = 2.

So the characteristic equation is t^2 - 2*sqrt(7)*t + 2 = 0.

Therefore, the recurrence relation for z_n is z_n = 2*sqrt(7)*z_{n-1} - 2*z_{n-2}

But since z_n is x^n + y^n, which is an integer? Wait, let's check for n=0,1,2.

z_0 = 2 (integer)

z_1 = 2*sqrt(7) (not integer)

z_2 = 24 (integer)

z_3 = 2*sqrt(7)*z_2 - 2*z_1 = 2*sqrt(7)*24 - 2*(2*sqrt(7)) = 48*sqrt(7) - 4*sqrt(7) = 44*sqrt(7) (not integer)

z_4 = 2*sqrt(7)*z_3 - 2*z_2 = 2*sqrt(7)*(44*sqrt(7)) - 2*24 = 2*44*7 - 48 = 616 - 48 = 568 (integer)

z_5 = 2*sqrt(7)*z_4 - 2*z_3 = 2*sqrt(7)*568 - 2*(44*sqrt(7)) = 1136*sqrt(7) - 88*sqrt(7) = 1048*sqrt(7) (not integer)

z_6 = 2*sqrt(7)*z_5 - 2*z_4 = 2*sqrt(7)*(1048*sqrt(7)) - 2*568 = 2*1048*7 - 1136 = 14656 - 1136 = 13520

So z_6 = 13520. Therefore, (sqrt(7) + sqrt(5))^6 + (sqrt(7) - sqrt(5))^6 = 13520.

But we need (sqrt(7) + sqrt(5))^6. Let's compute (sqrt(7) - sqrt(5))^6. Since sqrt(7) > sqrt(5), sqrt(7) - sqrt(5) is a positive number less than 1. Let's check that.

sqrt(7) ≈ 2.6458, sqrt(5) ≈ 2.2361, so sqrt(7) - sqrt(5) ≈ 0.4097, which is less than 1. Therefore, (sqrt(7) - sqrt(5))^6 is a positive number less than 1. So (sqrt(7) + sqrt(5))^6 = 13520 - (sqrt(7) - sqrt(5))^6. Since (sqrt(7) - sqrt(5))^6 is less than 1, the greatest integer less than (sqrt(7) + sqrt(5))^6 is 13520 - 1 = 13519. Therefore, the answer should be 13519.

Wait, but let me verify this again. Let me check the calculations step by step.

First, x = sqrt(7) + sqrt(5), y = sqrt(7) - sqrt(5). Then x + y = 2*sqrt(7), xy = 2. Then the characteristic equation is t^2 - 2*sqrt(7)t + 2 = 0. Therefore, the recurrence is z_n = 2*sqrt(7)z_{n-1} - 2z_{n-2}.

So z_0 = 2, z_1 = 2*sqrt(7), z_2 = 24, z_3 = 2*sqrt(7)*24 - 2*z_1 = 48*sqrt(7) - 4*sqrt(7) = 44*sqrt(7), z_4 = 2*sqrt(7)*44*sqrt(7) - 2*z_2 = 2*44*7 - 2*24 = 616 - 48 = 568, z_5 = 2*sqrt(7)*568 - 2*44*sqrt(7) = 1136*sqrt(7) - 88*sqrt(7) = 1048*sqrt(7), z_6 = 2*sqrt(7)*1048*sqrt(7) - 2*568 = 2*1048*7 - 1136 = 14656 - 1136 = 13520. So that's correct.

Therefore, x^6 + y^6 = 13520. Since y = sqrt(7) - sqrt(5) ≈ 0.4097, then y^6 ≈ (0.4097)^6. Let's compute that approximately. 0.4^6 = 0.004096, but since 0.4097 is slightly larger than 0.4, maybe around 0.0043 or something. So y^6 is less than 0.01. Therefore, x^6 = 13520 - y^6 ≈ 13520 - 0.0043 ≈ 13519.9957. Therefore, the greatest integer less than x^6 is 13519. So the answer is 13519.

But wait, maybe my estimation of y^6 is too rough. Let me compute y^6 more accurately.

First, y = sqrt(7) - sqrt(5) ≈ 2.6458 - 2.2361 ≈ 0.4097. So y ≈ 0.4097.

Compute y^2 ≈ (0.4097)^2 ≈ 0.1679

y^3 ≈ 0.1679 * 0.4097 ≈ 0.0689

y^4 ≈ 0.0689 * 0.4097 ≈ 0.0283

y^5 ≈ 0.0283 * 0.4097 ≈ 0.0116

y^6 ≈ 0.0116 * 0.4097 ≈ 0.00476

So y^6 ≈ 0.00476. Therefore, x^6 = 13520 - 0.00476 ≈ 13519.99524. Therefore, the greatest integer less than x^6 is 13519, since 13519.99524 is just a bit less than 13520.

Therefore, the answer is 13519.

But let me check if there is another way to compute this, maybe expanding the binomial directly. Let's see.

Compute (sqrt(7) + sqrt(5))^6 using binomial theorem.

The expansion is:

C(6,0)*(sqrt(7))^6*(sqrt(5))^0 + C(6,1)*(sqrt(7))^5*(sqrt(5))^1 + C(6,2)*(sqrt(7))^4*(sqrt(5))^2 + C(6,3)*(sqrt(7))^3*(sqrt(5))^3 + C(6,4)*(sqrt(7))^2*(sqrt(5))^4 + C(6,5)*(sqrt(7))^1*(sqrt(5))^5 + C(6,6)*(sqrt(7))^0*(sqrt(5))^6.

Compute each term:

C(6,0) = 1, term = 1*(7^(3))*(1) = 343

C(6,1) = 6, term = 6*(7^(2.5))*(5^(0.5)) = 6*(sqrt(7^5)*sqrt(5)) = 6*sqrt(7^5 *5). Wait, this seems complicated. Maybe it's better to compute numerically.

But since we need to find the integer part, maybe this approach isn't the best. Alternatively, perhaps recognizing that the method with the conjugate gives an exact integer for x^6 + y^6, and since y^6 is less than 1, x^6 is just less than that integer. Therefore, floor(x^6) = (x^6 + y^6) - 1 = 13520 - 1 = 13519. So that's the answer.

Therefore, I think the answer is 13519. Let me just confirm once again.

Yes, using the conjugate pair method, we found that x^6 + y^6 = 13520, and y^6 is a positive number less than 1, so x^6 must be 13520 - y^6, which is less than 13520. Therefore, the floor of x^6 is 13520 - 1 = 13519. So the greatest integer less than (sqrt(7) + sqrt(5))^6 is 13519.

**Final Answer**
\boxed{13519}
Therefore, the greatest integer less than $\left(\sqrt{7} + \sqrt{5}\right)^6$ is $\boxed{13519}$.
</think>