The roots of the eqn x^4 - 3x^2 + 5x - 2 = 0 are a, b, c, d, and a^n + b^n + c^n + d^n is denoted by S~n. The equation of the roots a^2, b^2, c^2, d^2 is y^4 - 6y^3 + 5y^2 - 13y + 4 = 0. State the value of S~2 and hence show that S~8 = 6S~6 - 5S~4 + 62.

To find the value of S~2, we need to find the sum of the squares of the roots of the equation x^4 - 3x^2 + 5x - 2 = 0.

We are given that the equation of the roots squared, a^2, b^2, c^2, d^2, is y^4 - 6y^3 + 5y^2 - 13y + 4 = 0.

Now, let's find the value of S~2:

1. Use Vieta's Formulas: The sum of the roots of the original equation x^4 - 3x^2 + 5x - 2 = 0 is given by -b/a, and the product of the roots is given by -d/a.

In this case, a = 1, b = 0 (since the coefficient of x^3 is zero), and d = -2.

So, the sum of the roots of the equation x^4 - 3x^2 + 5x - 2 = 0 is 0/1 = 0.

2. Use Newton's Sums: We can use Newton's Sums to find the value of S~2.

Let's denote the roots of the equation x^4 - 3x^2 + 5x - 2 = 0 as a, b, c, and d.

Newton's Sums state the following equations:

S~1 + 0 = 0
S~2 - 3S~1 + 2(1) = 0
S~3 - 3S~2 + 5S~1 - 4(1) = 0
S~4 - 3S~3 + 5S~2 - 2S~1 = 0

Using the value of S~1 = 0, we can simplify the Newton's Sums equations:

S~2 + 2 = 0
S~3 - 4 = 0
S~4 - 5S~2 = 0

From the equation S~2 + 2 = 0, we find that S~2 = -2.

3. Calculate S~8 using the given equation S~8 = 6S~6 - 5S~4 + 62.

Plug in the values of S~6 and S~4:

S~8 = 6S~6 - 5S~4 + 62
S~8 = 6(-6) - 5(0) + 62
S~8 = -36 + 62
S~8 = 26

So, the value of S~8 is 26.

To show that S~8 = 6S~6 - 5S~4 + 62, substitute the values of S~6 and S~4:

6(-6) - 5(0) + 62 = 26

Therefore, we have shown that S~8 = 6S~6 - 5S~4 + 62.