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answer:To find the sum of the squares of the solutions, we first need to find all the roots of the equation. We are given that two of the roots are 2 ± √2. We can use this information to factor the given polynomial. Let's start by rewriting the given polynomial as x^4 - 4x^2 - 8x + 4 = (x^2 - (2 + √2)x)(x^2 - (2 - √2)x) + Ax^2 + Bx + C, where A, B, and C are constants to be determined. However, this approach can be quite cumbersome. A more elegant approach involves using the fact that the given polynomial has two roots of the form 2 ± √2. This indicates that the polynomial might be a quadratic in disguise, with x^2 taking the place of the variable. We can rewrite the given polynomial as (x^2)^2 - 4(x^2) - 8x + 4 = 0. Rearranging the terms, we have (x^2)^2 - 4(x^2) + 4 = 8x, which looks like a quadratic equation in x^2. Notice that (x^2)^2 - 4(x^2) + 4 = (x^2 - 2)^2. Thus, we have (x^2 - 2)^2 = 8x. Rearranging the terms, we have (x^2 - 2)^2 - 8x = (x^2 - 2)^2 - 8x + 4 - 4 = (x^2 - 2x - 2)(x^2 + 2x - 2) = 0. The roots of the first factor, x^2 - 2x - 2 = 0, are indeed 2 ± √2, as given in the problem statement. The roots of the second factor, x^2 + 2x - 2 = 0, can be found using the quadratic formula and are -1 ± √3. We now have all four roots of the original equation. To find the sum of the squares of these solutions, we can use the fact that the sum of the squares of the roots of a polynomial is equal to the negative of the coefficient of the second-highest degree term, divided by the leading coefficient. In this case, however, it's simpler to compute the sum directly. The sum of the squares of the roots is (2 + √2)^2 + (2 - √2)^2 + (-1 + √3)^2 + (-1 - √3)^2. Recall that (a + b)^2 + (a - b)^2 = 2a^2 + 2b^2 for any a and b. Applying this formula, we have (2 + √2)^2 + (2 - √2)^2 = 2(2)^2 + 2(√2)^2 = 8 + 4 = 12, and (-1 + √3)^2 + (-1 - √3)^2 = 2(-1)^2 + 2(√3)^2 = 2 + 6 = 8. Therefore, the sum of the squares of the solutions to the given equation is 12 + 8 = 20.

answer:We have already found the roots of the given equation: 2 ± √2 and -1 ± √3. To find the sum of the cubes of these solutions, we can use the fact that the sum of the cubes of the roots of a polynomial can be expressed in terms of the coefficients of the polynomial and the sums and products of the roots. However, it's simpler to compute the sum directly. The sum of the cubes of the roots is (2 + √2)^3 + (2 - √2)^3 + (-1 + √3)^3 + (-1 - √3)^3. Recall that (a + b)^3 + (a - b)^3 = 2a^3 + 6ab^2 for any a and b. Applying this formula, we have (2 + √2)^3 + (2 - √2)^3 = 2(2)^3 + 6(2)(√2)^2 = 16 + 24 = 40, and (-1 + √3)^3 + (-1 - √3)^3 = 2(-1)^3 + 6(-1)(√3)^2 = -2 - 18 = -20. Therefore, the sum of the cubes of the solutions to the given equation is 40 - 20 = 20. Notice that the sum of the cubes is equal to the sum of the squares we found earlier, which is an interesting coincidence. This is likely due to the specific structure of the polynomial and its roots, rather than a general property of polynomials.

answer:We have already found the roots of the given equation: 2 ± √2 and -1 ± √3. To find the sum of the 4th powers of these solutions, we can use the fact that the sum of the 4th powers of the roots of a polynomial can be expressed in terms of the coefficients of the polynomial and the sums and products of the roots. However, it's simpler to compute the sum directly. The sum of the 4th powers of the roots is (2 + √2)^4 + (2 - √2)^4 + (-1 + √3)^4 + (-1 - √3)^4. Recall that (a + b)^4 + (a - b)^4 = 2a^4 + 12a^2b^2 + 2b^4 for any a and b. Applying this formula, we have (2 + √2)^4 + (2 - √2)^4 = 2(2)^4 + 12(2)^2(√2)^2 + 2(√2)^4 = 32 + 96 + 8 = 136, and (-1 + √3)^4 + (-1 - √3)^4 = 2(-1)^4 + 12(-1)^2(√3)^2 + 2(√3)^4 = 2 + 108 + 54 = 164. Therefore, the sum of the 4th powers of the solutions to the given equation is 136 + 164 = 300, but we must recall that there are other relationships between (a+b)^4 + (a-b)^4 and (a^2+b^2)^2 - thus another way to reach this answer can be found. (2 + √2)^4 + (2 - √2)^4 + (-1 + √3)^4 + (-1 - √3)^4 can also be expressed using the values of (2 + √2)^2 + (2 - √2)^2 + (-1 + √3)^2 + (-1 - √3)^2. Using the previously found solution to (2 + √2)^2 + (2 - √2)^2 + (-1 + √3)^2 + (-1 - √3)^2 = 20, we can then square this number to find ((2 + √2)^2 + (2 - √2)^2 + (-1 + √3)^2 + (-1 - √3)^2)^2.